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An alyti cal Trigonome try R_elationships between the trigonometric functions, known as trigonometric identitieso play a vital role in ca-lculus. Also, an alternative to the rectangular coordinate system, called the polar coordinate system, often is better suited for certain curves. Both of these concepts occur in the following example.* Find the points of intersection of r _ cos 20 and r - cos g, for0<0 a 2n. Solution Set cos 20 - : 2 cosz 0 - l, we get 2cosz0_ l:cosg 2cosz e - cos 0 - | - 0 (cos0-l)(2cos0+1):0 cos 0. Using the identity cos 20 Thus, l cos0_ and cose_ L -2 Hence, 0 0, 2n13, 4n13, so that the points of intersection are (- U2,2nl3), (1, 0), To check the pole, we set r For r - cos 20 with r 0, - cos 20 Forr-cosgwithrcos 0 :0, - and (- U2, 4nl3). 0 in each equation: so 0 :1 4 Pole is a point on the graph 0 - 0, so 0 -L2 Pole is a point on the graph The identity cos 20 _ 2 cosz 0 _ I is one of many we will encounter in this chapter. The polar graph of r : cos 20 is calle d a teaf rose, and the polar four graph of r_ cos g is a circle. These are illustrated in the figure. Note also that the solution of a trigonometric equation is involved in the example. All of these are things we will study. * Abe Mizrahi and Michaer Sulrivan , carcurus and Geometry (Belmont, Calif.: Wadsworth, Publishing co., @ L9g2 wadsworth, Inc.), p. -Analytic 657. Reprinted by permission. 309 310 1 B/ANALYTTcAL rRrGoNoMErRY Trigonometric Functions of Real llumbers So far our definitions of the trigonometric functions have had as domain the set of all admissible angles. The values depend only on the angle, and not on whether the angle is measured in degrees or radians. The next stage of this course is an important one, although it is somewhat subtle. We wish to gire meaning to such expressions as sin x, cos x, and tan x, where x stands for a reai number. Toward this end, let us recall that in the definition of the radian measure of an angle, we used an arbitrary circle with its center at the vertex of the angle. We find it useful in the present case to specify a circle of radius I , with center at the origin, and we refer to this as the unit circle (Figure 1). If we are given an rtt" angle of 0 radians, then the arc length subtended on a circle ofradius r is s 1, the arc length s equals the angle 0. In other words, the linear measure so if r (using the measurement units of r) of the arc length equals the angular measure : : of the central angle in radians. This relationship provides us with a way of defining the trigonometric functions of real numbers that is consistent with our previous definition. Figure 1 a A, 1 \ x + Figure 2 0 Given any real number 0 (not at present to be thought of as the measure of an angle, but just a number), we mark offa distance of 0 units on the unit circle. starting from (1, 0) and moving counterclockwise when 0 is positive and clockwise when 0 is negative. Let P(0) be the point arrived at in this manner with coordinates (x, y), as shown in Figure 2.* With this established, we can define the following: * P(0) is thus a function of the real variable 0, whose range consists of ordered pairs (x,.rr representing points on the circle. m fFl IiTRIGONOMETRIC FUNCTTO\S OF RE{L fu EiM D EFII\ITION sin0 1 1 -y fiEffi.r0M mff'ffihmrl cOS0:x [$ gmwr CSC0-v (r'* sec0-1 x @*01 0t l' &,,ttiumu tan o ilUWG -Lx (x +0) cot o:! (l*o; v 'iilfiir' ryurwlitu, m@"wl b,@,M frffisMmrH The relationship between the functions defined in terms of numbers and those defined in terms of angles should now be apparent. we simply consider the angle with initial side the positive x axis and terminal side connecting the origin and P(0), as shown in Figure 3. Then the radian measure of this angle is precisily 0, the same as the number of units of length of arc. Now by the definitions of the trigonometric functions of the angle 0 given in Section 4 of chapter 7, we Fffifrrmtlmrfifltr m'y Note that for the tangent and secant, 0 cannot be any number for which and for the cotangent and cosecant, 0 cannot be any number for which :0; :0, ! x r:,idit utrtffi: Ufoi,gnruU," have: sin 0 cos 0 tan 0 yy :y r1 xx r1 -x v x csc0:T v sec 0 cot ofi' wmfu kdfi*lruP r1 I for the functions of the number 0, we see that they are precisely the same. Suppose, then, that we are given the expression o'sin 2." How shall we interpret it? Does it mean the sine of the number 2 or the sine of the angle whose radian measure is 2? The answer is that it does not matter, because the values are the same. conceptually these are quite different, but sin 2 where 2 is a number has the same value as sin 2 where 2 is the radian measure of an angle. More generally, any trigonometric function of a real number has the same value as the same trigonometric function of the angle of g radians. Thus, insofar as M,&#nmch- v 0:!v comparing these with Definition [rc 1 Figure 3 units 0 radians 312 8/ANALYTICAL rRrcoNoMErRY values of the trigonometric functions are concerned, it makes no differene whether we think in terms of an angle measured in radians or in terms of thr real number that equals the number of radians in the angle. EXAMPLE 1 x coordinate of the point P(0) on the unit circle is and P(0) is in the second quadrant. Find the y coordinate of P(0) and write all six trigonometric -!, The functions of0. Solution By the Pythagorean theorem we have (-+)'+v2:1 16 ,,2 1 J12525 - 9 y -* By Definition 7,, sin 0 '5 at EXAMPLE 2 csc 0 4 x- -t :L : 3 cos0_ tan 0 3 v 1 5 v 3 1 sec 0 x 5 -4 4 -x - v-_ -;J cot 0 4 Determine the value of e, where 0 < 0 <2n. P(0) (-,fr12, -+) b. P(0) (- a. Solution b. - - rlo,uo) We observe that the reference angle of the angle determined by the radius drawn to the given point is 30", or 7116 radians. So the angle itself is 7n 6 radians. Therefore 0 :7n|6. The reference angle in this case is 45", or nf4 radians, and the angle is therefore 3nl4 radians. So 0 :3n14. Figure 4 Figure 5 I/TRIGONOMETRIC FUNCTIO\S OF REAL qE@ \L\TBERS 313 If you are using a calculator, you can find trigonometric functions ol real numbers by considering the number as the radian measure of an angle- and this can be done regardless of the size of the number. For example. using a cale-utator we can redd trrfu sin 2 - cos 10 0.9093 : -0.8391 tan(-3.5; - -0.3-46 aqd so on. However, to read these values from a table such as Table \- ar the end of this book requires so{ne preliminary steps. This is because the table onllgives functions of angles between 0' and 90o, o1 equivaleptly, between 0 and nf2 radians. For larger angles, we make use of reference angles. Since z 2 = 1.5708, we cannot find'functions of numbers grepter than this directly from the table. To get sin 2, for example, we proceed as follows: As ap angle in radians j hes in the second quadrant (between nl2 x 1.5708 and n x 3.1416). We shorv this in Figure 6. The refergnce angle is thereforg n - 2 x 1.1416. Since the sine is positive in the second quadrant, we have that sin 2 is approxirnately equal to sin 1.1416. Reading from Table V (and intelpolating between values found there), we see that this is approximately 0.9093. We could also obtain cqs 2 by observing that the cosine is negative in quadrant II.'Thqn, again usipg the reference angle, we find cos 2 x -0.4l6L Thus, the coordinates of the point P(2) on the unit circle are approximately (-0.4161, 0.9093). Figure 6 P(2) r ( - 0.4161, 0.9093) : : tan2 sin 2 0.9093 coS 2 -0.4161 -2.185 To find the functions of 10 using Table V, we again must find the reference angle. Since 10 exceeds 2n,the first thing to do is to subtract 2nftom 10, in order to obtain an angle in the first revolutipn that is coterminal with 10. This gives l0 - 2n nv 10 - 6.2832:3.7168 This is seen to be in the'third quadrant (Figure 7), and so the reference angle is 3.7t68 - 3.1416 :0.5752 From the table we can flow obtain sin 0.5752 : 0.5440 and cos 0.5752: 0.8391. Since both sine and cosine are negative in the third quadrant, we have finally that sin 10 t -0.5440 and cos 10 = *0.8391 The advantage of a calcplator for finding functions of real numbers should be evident. 314 8/ANALYTTcAL TRrGoNoMErRy Figure 7 x P(10) -P(10-2n) ! (- 0.8391, EXERCISE SET A 3.7169 _ 0.5440) 1 In Problems 1-10 flnd all six trigonometric functions of r. p(0): (-+, 3' P(0): G, -P.) -zrfztzl 0. z. rp1:1_z1ls, U\fs) 4. P(0)= (-0.6,0.8) 5. The x coordinate of P(0) is f, and p(0) is in the fourth quadrant. 6. The y coordinate of P(0) is f , and rl2 < 0 < n. 7. The abscissa of P(0) is -1, and n < 0 < 2n.Whatis the value of 0? 8. The ordinate of P(0) is uE1z, and the abscissa of p(0) is negative. what is the value of 0? 9. 10. The abscissa of P(0) is {, and the ordinate is negative. The ordinate of p(0) is and 3n < 0 < 4n. _*, ll and 12 give the coordinates of the point p(0) for the given values of 0. ll. a. P(nl2l b. Plrl c. P(3n12) d. P(5n16) e. P(anl3) 12. 1 P(snl4) b. P(8n13) c. p(-7nla) d. P(5n) e. P(3lnl6) In Problems ln Problems 13 and 14 indicate the approximate lgcation of p(0) on the unit circle for tbe given value of0. 13. a. 0:0.5 14. a. 0:8.2 In Problems 15 and b. e--2 b. 0:L2.75 16 give the value 15. a. P(0) 61 312- r) c. P@) -- (+, ,lZlz) e. P(0) - (1, 0) 16. a. p@) - ( - ,,_- ,ll tZt c. P(0) - (- J 312, +) e. P(0) - (- 1, 0) In Problems 17 and 18 c. 0:4 c. e--I.54 d. e- -15 d. 0-4.72 of 0 if 0 < 0 <2n. b. P@)= (1 IrE,rlJ2) d. P(q -. (0, 1) b. P(0) _ (0,_- 1) d. P(0): (J212, _rlZlz) flnd sin 0, cos 0, and tan 0 using either a calculator or tables. 17, a. 0 : 0.3 b. 0 : 2.718 c. 18. a. 0 : 7 .1432 b. 0 : 0.2967 c. 0 0 - t4.03 d. e - -3.625 _-5.1324 d. 0:10.5146 2/SOME BASIC TRIGONOMETRIC 19. Find all values of 0 such that 0 a, sin0:l d. sin0-0 g. sin0--l B 20. Use the 21. < 0 < 2n for which: b. cos0-l e. cos0-0 h. cos0:_1 c. tanl:l f. tan0_0 i, tan0__l unit circle to show that: n. sin(zu - 0) _ sin 0 c. sin(z + 0) _ - sin g e. tan(n+q_tan0 b. d. IDENTITIES 315 cos(zu cos(zr - g) : - + 0) : - cos g cos 0 Show that for all integers n, each of the following is undeflned for the stated values of 0: a. tan0 for 0- 2n*l b. cot? for 0:/ur -,-n c. sec0 for 0- -2n*1 2 " 22. Show thar lsec 0l > 23, Prove that the range 24. Let 1 .s- d. csc0 for 0:fin and ncsc 0i > t for all 0 in the domains of these functions. of tan 0 is the set of all real numbers. {rr^ : t T - {0 e R: U-{keR: ^ (2n * ltn U+ ,il:0, rl, -2, Z " 0+nn,n:0, +1, +2, ,JI lr'l ] >U Prove that if f(0): Sec 0 and g(0): csc 0, then: a. "f is a function from S onto'U b. g is a function from T onto U Is either f or g a l-1 function? Explain. 7 Ll Some Basic Trigonometric Identities You may have already noticed the following relationships among the trigonometric functions: tano-sino cos 0 cotU- cos 0 sin 0 SeCU-- 1 cos 0 These are seen to be true from Definit ion 2 of chapter 7 or Definition I of this chapter . They are examples trigonometric identities-so-called because they -of are true for all admissible values of a. There are many other identities, and sorrie 316 8/ANALYTTcAL TRTGoNoMETRY are so basic that they should tje committed to mbmory. These will be indicated as we proceed. The Pythagorean theorem is the basis for another important set of identities. Using the theorem, we have, for (x, y) on thb unit circle (see Figure 8), x2+Y2:1 Figure 8 But x - cos 0 and y - sin 0. So tn2 0 If we divide both * cos2 0 - 1 sides of the flrst equation by x2 (assuming x * 0), we obtain which, according to our definitions, is 1*tan20-sec20 However. if we divide the first equation by y2 , we get (r'*1 :6)' or co( o +1 - cscz o We collect these results together: sin2 0 + cosz 0 t + tan2 0 - 1 sec2 g 1+cot20-csc20 EXAMPTT These are known as the Pythagorean identities and should be learned. Notice the similarity between the last two. In applying these, it may be necessary to write them in various equivalent forms, for example, cos, I sin2 6 or sec2 0 | tanz 0, but it is probably besi to concentrate on learning them in - : b: - 2/SOME BASIC TRIGO\O\IETRNC Ibe indicated IDE\TITNES 317 just one form, such as the one given. you can then mentallr renrire then: in various ways. Starting from any point on the unit circre, if we go 2zr unirs along rhe cir;lc either clockwise or counterclockwise, we wind up ut th. same poini. srnce rhe circumference of the unit circle is 22. Thus, fo. any 0, pldl: prd 3:-. . where n is any integer, positive or negative. Since the vaiues of the trigonomeri:; functions depend only on the coordinates of p(0), it follows that cf identities. rre 8), : sin(0 * cos(0 + 2nn): 2nn) si^ g sss 6 and so on, for the remaining trigonometric functions. These can be expressaJ briefly by saying that each of the trigonometric functions is periodic, rvith period 22.* Figure 9 v P(0\ \ -(r-v (1, 0) t// h. n-e obtain P(- t )-(x x -v) consider next the relationships between the functions of 0 and -0. By symmetry we see that if the coordinates of p(0) arc (x,y), then those of p( 0; (x, -y) (Figure 9). The sine of g is by deflnition the coordinate of p( - g) 7re 7 but this is the negative of the y coordinate of g. Thus, sin( _ 0) : _ sin 0. Since the x coordinates of P(0) and p(-0) are the same, it follows that cos(-0): cos 0. similar relationships for the other functions can now be obtained: -0): -sin g cos(- 0) - cos 0 tan(- 0) : - tan 0 sin( -0): sec( - 0) csc( -csc 0 sec 0 cot(-0): -eote : It : is most important to learn the relationships sin( 0) - sin 0 and cos( _ gt cos 0, since they occur more frequently than the others and since the others can be deduced easily from these two. EXAMPLE ned. Notise rcessary to sin2 g or img them in - 3 Make use of the basic identities of this section to find the other fir.e trisonometric functions of 0 if tan 0 : -2 and cos 0 < 0. : + In gen^eral, if /lx + k) f(x) forall x, then is said to be periodic with period k The smallest positive fr for which this is true is called the rr"i"."nta period, For the sine, cosine. secant. and cosecant, the fundamental period is 22, but for the tangeni and cotangent, / ;ti, ;--' 318 8/ANALYTTcAL TRTGoNoMETRY Solution Firstwe observethatcot 0 : lltan0 : -L.Next, since sec2 0 : | + tanz 0 : I + e42: 5, we have sec e : fJ5. But cos g is negative, so that sec 0. which is 1/cos 0, is also negative. Thus, sec e: -J5. Therefore, cos 0: _rtJs. : - : : - Nowweuse sin2 0 | cos2 0 I + f. Sincetan 0 andcos 0 areboth negative, and tan 0 sin 0f cos 0, it follows that sin 0 is positive. Thus, sin 0 2l'f 5. Finally, csc 0 l/sin e {512. : : : : Note. We could also have determined the signs of each of the functions b1' observing that for both tan 0, which is y f x, and cos 0, which is x, to be negative. we must have y positive, so that the point P(0) is in the second quadrant, or equivalently, the angle of 0 radians terminates in the second quadrant. EXAMPLE 4 Solution In each of the following use the basic identities to show that the first expression can be transformed into the second: a. sin 0 coso tan U'" a. Since tan 0 - cos 0 b. --- ^" csc0- sin0 tan U' sin 0/cos 0, we have sin 0 sin 0 cos 0 : ,-sin0._-cos0 slnU tanA smU C"S O cos0 ^ cos0 : cos ' t^" e ' ,i., o b. cos2g ,ir, o - 1 - sin2 0 andcsc g : l/sin 0 to get cos2 g 1 srnz0 : 1-sin20:rirr0s", 0 ,irr0 U"0-csc0- sin 0 Now we use the identities cosz 0 By using the basic identities we can proye a multitude of others, but fortunately it is not particularly useful to try to memorize any of the results. B5proving an identity, we mean showing that the given equation is true for all admissible values of the variable or variables involved. The general procedure is to work only on one side of the equation and, by use of the basic identities. transform it so that in the final stage it is identical to the other side. Often the more complicated side is the better to work with, since it offers more obvious possibilities for alteration, but there are times in calculus when it is better to change a simpler expression into a more complicated one. We illustrate the procedure by means of several examples. EXAMPLE 5 Solution Prove the identity tan 0 + cot 0 - sec 0 csc 0. Perhaps it would be better to word the instructions, "Prove that the following is an identity." For we cannot prove it is true by assuming it is true, and this is an important point of logic often missed. This precludes, for example, working E\A}IPI 2/SOME BASIC TRIGO\O\IETRIC -{l- t-m] # ,: thr re' on both sides of the equation, unless we verify that each step is rerersible. A proper approach is to begin with the left-hand side and try to obtain the rigtrthand side: ,rru d&" $cr$ f- tano+coto:sino*coso r # are thrmfu [m- su:: * IDE\T[T[ES 319 : cos 0 sin 0 sin20+cos20 mtiors cos 0 sin 0 mS henrymru,e- uadran:* 1 cos 0 sin 0 $m ntnL ll :"or0'rirr0 I enqwesisnilcmml and now the given.equation is verified. It is true for all admissible values of 0, and in this case this means all values except 0, nf2,3nl2, and any angles coterminal with these since, at each of these values, two of the given functions are undefined. usually we will understand that such exceptions are necessary without mentioning them explicitly. EXAMPLE 6 Prove the identity 1 sec, + r#to ger 1 : cotx csc Jr - csc2 x+ 1 Solution It would be possible to transform the right-hand side into the left, and may even be easier, but we choose to work from the left-hand side because in later applications the right-hand side will be seen to be the more desirable final form. We begin with a commonly used trick, that of multiplying numerator and denominator by the same factor. ; but frorrnnrmuhs" mB trffi flcr effi I pooedrrru r idemtiuifis" r. Oftem thc srg obr-iloum ir better ts hstrate secx+1 1 secx-1 secx-l secx+1 secx-1 sec2x- 1 The object here is to bring sec2 x 1 into the picture, because this is, by one of the Pythagorean identities, equal to tan2 x. So, 1 :...,-secx-1 secx+1 tanzx cos2 x cotz x x sin2 x cos x (csc2 x - 1) -:Tsln' x - COSX and fihis te m orking tanz x cos : ncn 1 1 tk n foflomring secx tanz x sin x 1 sin x -cotxcsc x- csc2 x+ 1 320 8/ANALYTICAL TRIGoNoMETRY EXAMPLE 7 EXE Prove that 2 sin3 x 1-cosx 2 sin3 x 1-cosx Solution EXAMPLE 8 Solution 2 sin x(1 * cos x) 2 sin x(1 - cos2 x) 1-cosx 2 sin x(1 * cos xX1 1-cosx - 2 sin x(1 * cos x) cos x) Pror,e the identity csc0-cot0 :Cot0 sec0-1 1 cos0 *r0-rir0 csc0-cot0 sec0- | 1 1 cos 0 1-cos0 rir, 0 1-cos0 cos 0 :- 1-cos0 sin 0 cos 0 1-cos0 cos 0 sin 0 :COt0 A word is in order about the utility of proving identities. In calculus, as well with an unwieldl expression involving trigonometric functions. Often, by use of the basic identities, these can be transformed into more manageable forms. So in practie we usually do not have a ready-made identity to verify, but rather an expression that is to be changed to some other, initially unknown, form. It would be more accurate to describe this procedure as deriving an identity. The value of proving identities already provided lies in the fact that it gives a goal tc-r as in more advanced courses, we are frequently confronted shoot for in transforming an expression. Without this and with little or no experience in such activity, a student may go in circles or change to a less desirable form. R IDE\TrrrES 321 2/SOME BASIC TRIGONON,{ETRTC EXERCISE SET A In Problems 1-8 find the rbmaining five trigonometric iunctions of 0. usine the basic identities of this section. 1: 3. 5. 7. sin0:-!,' tang>0 tan0: -4, nl2<0<inl2 cos?:2lJi, cot0<0 sin0:*, 5nl2<0<7n12 i. r."b:g, cscg<0 4. eotl:#, n<0<2r 6. csc0--+, tan0>0 8. secg:3, 3n<0<4x In Problems 9-18 show that the first eipression can be transformed into the secon,J. using basic identities. 9. tan 0 10. sec0 SIN H- -, sin 0 11. ----- . sec 0 cot e' 13. cot 0 sec 0, 15. SCC O , - cos 0 csc 0 11 - -. 17. --- nr sec'0 cscz cos 0 12. sec2 14. tan 0 csc 18. I 0' . 0' csc 0 stn2 0, g, sin 0 16. 1- tan 0 CSC O' cot 0 _. sec2 0 sec 0 sin2 0 sec2 - - l cosz0 0' csc sec2 0 0, 1 In Problems lg -67 prove that the given equation is an identity by transforming the leit-hand side into thi right-hand si&. 19. 1+sin0 :cos?+cot? 20. sinxcotx-sinx zl. ] zz. 2i. 24. (tan tan 0 COSXCSCX - sec2 o xtanfi:cosx (1 + co* o) tan2 o 25. secx-sin sec0-cos0 --sin0 tan 0 sin2 x-l , -:CSC2J-1 cos'x-I x+cotx)2:sec2 x+ csc2x tan 0 cos 0 26. 1--:COSZ0 CSC O - : L- 27. (sin 29. sin0tan0:sec0-cos0 31. (tan2 0 33. 35. 37. @ cos 0)' + lxcosz sina x-cos4 0 2 sin @ - 1) - 1 - cos d sec2 0 ff-2sin2x-1 tan0+l ^--sin0 sec0*csc0 sec2 x 39. j I*cot'x sin 0 32. csc 0 34. 38. 41. (1 * -tan2 x csc 0)(sec 0 - tan 0) 40. : cot g sec2 a 30. 1 .-sinOcos0 - 1 . cos- d" - | -sec2z 36. tan0*cot0 1- 28. 42. * - cos 0 cot 0 cos 0 cot 0 - csc 0 sin 0 (1 - tan 0)2 \;z,:-1-2sin0cos0 cos x cot x -==---- 1csc x cos 0(tan e - -sin2 x sec 0) -sin0 1-csc0 secx*tanx _ coS x(1 + sin -rl -, SEC- X 1-cosx sinx Stnx +-_?csc.r 1-cosr 322 8/ANALyrrcAL TRrcoNoMErRy 43, tan2x- sin2 tan2x-l a45. :: sln' x 47. sec4 . SeC2 ft:tan2 L+tanx : 49. ;I*cotx 11, : :2sec2-r 1-sinx -: 1+sinx 46. cot4 A - l: cot2 g csc2 0 - cscz 0 ff:tanz xsin2x x - 44. x csc2 t^ tanz a 48. xsec2 x* sec2x t€Ifl sec- r 1 SeC x *tanx - tanx sinx L -cosx 53. ;--.--:2cotx I_COSX SlNX oo...? -: r1 . - oww- X -- 51. 55. sec x ;---; 1 - csc'x - 56. (cot g B csc tan2 0)' ' .rr sec2 :1 - .\ - sec2 .y + d. sin2 a - I - sin: z sin x 50. cotx*cscx -l-cosx 52. tana s4. $an2 ft- sec2 xtanz x-sec2 o-- l) co: o: sin0-cos0 sec o + x*l csc o 1 cos 0 l*cos0 57. (1 - cos2 0)(2 * tanz 0) : sec2 0 - cosz 0 58. tan g(sin 0 + cos 0)2 + (1 - sec2 g) cot 0 :2 sinz 0 59. 2 cos x - sec x(l - 2 sin2 x): sec x sin 0 60. ;-;:cSC0+cot0 I-cosU 62. 1-sinx , . ,tr, - 63. -1rsin0 *sec20-tangsec0+l M. 66. 67. 68. t 2 sec2 x - 61. 1 , 1-sin0 sec2 0 + tan g scc g -: 2 sec x tan x - ! sin 0 sin: .r tanl .x . :toox*s1nx tanr-sln.r sin3 0 - sins 0 cos 0 sin 6 - sec0-cos0+tan0 -sin0 tan0*sec0 65. sin g cos2 g + cos3 0 -r- cos 6 - (sin 0 - cos 0)2 /i -cosd sin 6 l0<0<n) , l+cosU V1+cost,- _, By making the substitution t:2 sin g, where -trl2 < 0 < 7rlz, show that t_ t ------: "rl4 - t2 69. By making the substitution t - tan 0, where JTTT: - Tl2 < 0 < nl2, show that csc o t 70. Iftan0: tan 0 t, prove that singcos O:-J- 71. Bymaking the substitutio nt:*sec0, where'o*.1'r.nl2if if < o' show that ' ^trF t - 9 t>0 and z <0 <3n,2 3/THE ADDTTTON FORMULAS FOR SrNE AND COSTNE 72. Make the substitution: x : sin aQnp < 0 < nl2) in the 73. 74. 75. x21{t *} and show the result can be written in the form sec 0 - cos 0, Let x : 2 tan 0(-nl2 < 0 < nl2) in xll[ a], and.show the result is sin 0. Substitute x : 3 sec 0, where 0 lies either in the first or third quadrants, in (x2 * 913t2 1x, and show the result can be written in the form 9(sec 0 tan 0 - sin 0). Substitute I : I tan 0(-n'2 < 0 <nl2) in xl1@p a 9 and show that the result is 1sin 3 expression 323 0. The Addition Formulas for Sine and Cosine We will develop in this section formulas for the sum and difference of two numbers (or of two angles). These are of fundamental importance in deriving other identities. We designate by z and p two numbers which we will initially restrict to be between 0 and 2r and for which e > B. These restrictions will soon be removed. In Figure l0 we show typical positions of P(z) and P(p). Now the distance along the circle from (1, 0) to P(e) is a, and the distance to P(p) is B. So the arc length from P(B) to P(a) is a - B. If we think of sliding this latter arc along rhe circle until the point originally at P(fi coincides with (1, 0), then the distance a - p is in standard position and the end point of the arc is correctly labeled P(a - h. Since the x and y coordinates of a point P(0) are cos 0 and sin 0, respectively, we can show the coordinates of the three points P(a), P(p), and P(a - fi) as in Figure 11. Now, since the arc from (1,0) to P(u - fr) is equal in length to the arc from P(B) to P(a), it follows that the chords connecting these respective pairs of points are also equal. We can get these lengths using the distance formula: Distance from (1, 0) to P(u -b- Distance from P(p) to P(a) (cos a- cos p)'+ (sin a- sinB)2 Since these distances are the same, their squares are the same; so we obtain the equation [cos(a Figure 10 :3m;3 - P) - 1]2 + [sin(a - P)f' : (cos o( - cos p)2 + (sin a - sin B)2 324 8/ANALYTICAL TRIGoNoMETRY Figure 11 (cos B, sin p) P(p) - (cos(a 'P(a) p), sin(a - p)) (cos e, sin a) - On squaring and collecting terms, and making use several times of the identity sin2 0 + cos2 0 : 1, this gives cos2(a - frl -2 - cos(a P) +1 + sin2(a -B; : cos2 - 2 ccis a cos B * cos2 f + sin2 a - 2 sina sin B * sin2 B 2 - 2 cos(a - $ : 2 - 2lcosri cos B * sin a sin B) o( or finally, cos(a-f)=cosacosP+sindsinp (1) Now we remove the restrictions on o( and B. Suppose first that B > a, with u and P again in 10, 2n]. Then, since cos( - g) : coS g, we can apply (1) with a and p interchanged: cos(a - f) : cos(f - a) : + cos B cos a sin B sin a :coso(cosf+sinasinB So (1) still holds true. If a:p,it again holds, since the left-hand side is cos0, which equals 1, and the right-hand side is cos2 o( + sin2 a, which also equals 1. Finally, let r and fi be any two numbers whatever. Then we can write oc: a, * 2mn and, p : frr * 2nn, where m and n arc appropriately choben integers, and a, and f, lie in the interval l0,2rf. Then, by periodicity of the sine and cosine, cos(a -, : : :::i[l 1,, cos(ar - f r) r ::: :::: ],,:,,:;:ffo BY PeriodicitY i;ilTh"'' lI l'1,"."*, Thus equation (1) is true for all real numbers a and p. From equation (1) we are able to get d number of other results. Consider first cos(a + f). We write a + 0: u - (-0 and apply (1) with p replaced bv -fr: cos(a + P) - cos[a- (-p)]: coso(cos (-fr) * sinasin(-fr) Since cos( - P): cos B and sin( cos(a - fr): -sin B, this gives + p) - cos n cos p - sin oc sin B (2) 3ITHE ADDITION FORMULAS FOR SI\E {\D CCSI\E 325 Next, in equation (1) let n : nlz. Then, since cos n 2 - 0 amci sr-n a:- i. we get cos In this, put p - Tclz - (;-,) -sinf l_.tl e to get 'ln ,) -')] : sln\z l; - (; Simplirying, we see*; r) : ""(; - r ideriuil Now we can get sin(a $." srn: j sin(a [t) wryrr m-ilTh u fr): cosl;- : sin oc (-$ Ir 0) as follows: (a +r,] Bv (3) :cosl(;-.) -p] :.*(; - o).o, r + sin(; - -)',,, I >& + * cos g cos B + cos a sin p Bv (1) By (3) and () Notice that in applying equations (3) and (4) we have substituted different values for B and 0. Thus, ; fu ccqs ftcquals I vrite n i - imtegr,ru- fline effic sin(a * 0): sinacos B + cos a sinB - B : a + (- fr) and using (5), we get sin(a - B) : sin[a + (-0)] : sin a cos(-B) + cos d sin(-B) or, since cos(-fl : cos f and sin(-p) : -sind, sin(a - P): sinacos P - cos a sinB (5) Writing u (6) Formulas (l), (2), (5), and (6) are known as the addition formulas for the sine and cosine. They are of fundamental importance and should be memorized. We summarize them below. Notice that for the sine, the sign between Cmmdm rcpleffi{i u l,l,,r The Addition Formulas for Sine and Cosine + p) -sindcos fr+ cosasinB sin(a - p) -sinncos p- cosasinB cos(a + p) :coSflcos p- sinasinB cos(a - p) :coSflcos fr+ sinasinp sin(a 326 8/ANALYTICAL TRIGoNoMETRY the terms on the right agrees with the sign between a and f on the left, whereas for the cosine these are reversed. So there really are just two basic patterns to EXAMPL be learned. EXA MpLE 9 Solution Find the value of each of the following without the use of tables or a calculator a. sin(5nl12) b. sin(n/l2) c. cos(l3nll2) d. a. We can write 5nll2 as2nll2 : cos(-nll2) + 3nll2: nl6 + nl4. So .infi:,*(;.;) :.in1"o.1+"o.1ri,1 ---6---4 ---6 4 Now we know that we may treat of angles. Thus, sinsi=(;) #:""(# smr b. nf 6 and nl4 as if they are radian measures H-g) G) ::f t ?)-sin G : sin 1.o* Y * cos ! 4646 rinT f-r tr) G)0: 2J' G) C. 1,3n COS n?:cos tr.?):cos tr.oJ : cosf .o. ;- $,) d. cos( - n 1,2121 - cos (n rL2), since cos ( - 0) sin+sin 0 H€) =-1+.F ")L o\o : cos 0 ; so, co ",(- ?):cos #:cos tr-T):cos TE f TE TE G-t ) 7T :[l Hd;d) : .,F+ t zrfz EXERCN 3ITHE ADDITION F'ORMULAS FOR S[\E A\D COSI\E 327 n,yM.mm ntficrmm rum, ekufri&Irrr EXAMPLE If - rcl2 < x < nl2and 10 a. Solution y) b. y <7t, sin x 0< cos (x - y) - f,, andcos ) - -;. find: We conclude that x is between 0 and nl2, since if it were between 0 and l. sin x would be negative. Similarly, we must have y between nl2 and. z in order -;: l2I r sin(x + for cos/ to be negative. By the pythagorean identity, sin2 we get x + cos2.r: l. cos2x:l-sin2x So cosx- t@ The positive sign must be chosen, since 0 <x< 7T12. cosx- ffittlnlnurr$ Similarly, (-; We now have all the necess ary ingredients for obtaining the solutions. sin(x -t / +y)-sinxcos y+cosxsi ny: 43 ('_ :) -15 + \m 'fr b 24 b. cos(x - y): cos x cos y +sin x siny :{f-:) - 4 \i) :- -5\ '', ; % ..lllil . (+)$) t \-" ,r a '\y' - ) EXERCISE SET A 3 In Problems 1-8 evaluate without using a calculator or tables. l. 2. sin(a sin(a + P) and cos(a + P),where a - Tll|and fi - ni3 - P) and cos(a - P),where a - nl4and fr /5n z\ 3. a. ,in(!- I\ T) b' \r- 4. L.',,(?.T) 3 5. a. .o*tn '"'(?.;) b. ,",(+ r7n b. sin, -+) nl3 .1(#) + I ({ 7 + 3\, 24 11 328 8/ANALYTICAL TRIGoNoMETRY 6. a. / z\ r*(-r,/ b. / 5z\ *r(-ri b. cos 15' 7, L, sin 75" 8. a. cos 255' b. sin 195' 9. If sin a : -!, sin B: -*, P(a) is in the third quadrant, and P(p) is in the fourtL quadrant, flnd the following. a. sin(a - fl b. cos(a + p) 10. If cos a : {t, cos P : -1, P(a) is the fourth quadrant, and P(fl is in the second quadrant, flnd: a. B) b. cos(a - B) andsin p: -J, flnd: b. cos(a - B) a. sin(a + fl 12. If tanx: -|, cscx > 0,sec y : +,and coty < 0, flnd: a. sin(x - y) b. cos(x * y) 13. If sin a : -2l,,,B,tand> 0, cos B: -+, and csc B > 0, find: a. sin(a - fl b. cos(a + B) 14. lf cot A : -+, sec ,4 > 0, tan -B : t, and sin -B < 0, find a. sin(l + -B) b. cos(l -,8) 15. If sinx: -J, tanx> 0, secy: -+, and coty < 0, find: a. csc(x * /) b. sec(x - y) sin(a + 11. If sina<0, cos oa:$,cos p<0, : Establish the formulas in Problems 16 -20 by using addition formulas. b. cos(z+0)- -cos0 16. a. sin(z+0)--sin0 17. a. 18. a. ,,r(;.r-coso ,rr(*- o) : b. o b. .", (+ - g) : \2 ) 19. a. sin(e - 6) - sin 0 zo. a. rrrfi-r)-coso \2/\2/ cos ."'(;*r) :-sing sin o \2 / b. cos(n - A): -cos 0 b. ."r(i-r)-sing Prove the identities in Problems 21-29. 21. sin a cos p(cot a - tan $: cos(a + f) P _t ""t("-0-1+tarr.ta"P'r' sin(a*f)+sin(a-f) cos(a * p) + cos(a - p) sin(a * B) 26. :.: cot a + cot p sln o( srn p 28. sin(x * y) cos y - cos(x + y) sin y : 29. sin x sin(x * y) + cos x cos(x * .y) : .. "' cos(o(+ 0l n tand-tanf tandtan tuno+unB-sir4a+p1 - cos(a-B;-cos1a+p) * p) cos(a - B) sln sln l, sin(a 27. sin cos x l, o( -: B 30. Prove that, in general, sin(a * f) + sin a * sin B. 31. Find all values ofx for sin(a-0) which 0 < x < 2n and sin 5x cos 4x : cos 5x sin 4x - sin(a 1 - p) + cot acot B 4/DOUBLE-ANGLE, HALF-ANGLE. 32. Find all values 33. Derive a formuler for ofx for x: 1 - 2 sin 2,x sin -t a. sin(a+f+y) b. cos(a+f+y) Hint. Ure : r*;;1n REDLCTNO\ FI]R,\[I-'tr..{S 329 which 0 < x < 2n and 2cos2x cos in A\D First use the associative property for addition. Prove the identities in Problems 34-36. 34. sin(a * B) sih(a - 0) : .irr' a - sin2 B 35. cos(a * B) cos(a - f) : cos' a - sin2 B 36r sin(a * B) cos(a - f) : sin o( cos o( + sin B cos B im rhe ffi::mt[ : 4 Double-Angle, flalf-Angle, and Reduction Formulas The importance of the addition formulas in Section 3 lies primarily in the fact that so many other idehtities can be derived from them. we will carry out the derivations for some of the most important of these. The double-angle formulas are obtained from the addition formulas for sine and cosine of a and p by putting ll : a.If we denote the common value of a and p by 0, we obtain sin(, + 0):;:: sin 0 sin 0 sin 0 ;::: ;'o'g So, sin 20 Also, cos(O + 0) - 2 sin 0 cos 0 : cos 0 cos e : cos2 0 So, cos 20 - - sin2 0 = cos2 e - sin2 0 (7) Two other useful forms of cos 20 can be obtained by replacing, in turn, by 1 - sin2 g and sin2 0 by 1 - cos2 0. This gives cos2 0 ffil* and ffih cos20:(I -sin2 q- 0- 1 -2sin20 cos 20 : cos2 0 - (1 - cos2 0):2 cosz 0 sin2 1 :: lin'$m |liry]lr* rffir d rr So we have cos20=1-2sinz0 (8) | (e) and cos 20 = 2 cos2 e - whether to use (7), (8), or (9) depends on the objective. we some examples where a particular form is clearly preferable. If we solve (8) foi sin2 g and (9) foi cos2 0,we obtain sin2 0 - l-cos20 wiil shortry see 330 8/ANALYTICAL TRIGoNoMETRY and - = I +cos20 2 cos'0 These forms are employed extensively in calculus. By replaiing 0 by ul2 in the last two equations and then taking square roots. we get the half-angle formulas: ,ofr sm, rt- [1 d, cos,= *ai[+cosd z :{ -cosd 2 The ambigirity of sign has to be resolved in each particular instance according to the quadrant in which al?lies. From the addition formulas we cari obtain a class of identities sometimes called reduction formulas, of which (3) and (4) of Section 3 are special caser Here are some others: sin(rc + 0) tl,;;ot 0 + cos tr sin 0 .o, e _ 0 - sin 0 + = and sin(n _ q: cos z sin o :i, ; Similarly, cos(z and cos (n + 0) : .:X;' - r, .:X;' I '*(| and sin 0 sin z sin 0 these, we have used the facts that sin z si:n(3n12): - L and cos(3n12): 0, we also have In zz : 0 and cos fi : - 1' Sine * r) : rinf .o, o + cos !,in : -cos o ,) : ,i, f ,r"(!t : g "o, o - cos ! "i, o -cos0 Likewise, tL"r" ur" similar formulas for cos[(32/2) t 0]. To generalize, it appears that we should consider two cases: Case 1. Functions of (nn case 2. Functions * 0) j 6l@!!l * 6] L2 I [x 4IDOUBLE-ANGLE, HALF-ANGLE. AND REDLCTTO\ FCR\IL LqS 331 l where z is an arbitrary integer. Note that in case 2. an odd multiple of r nor*-be is involved since (2n + 1) is always odd. The following idsntitiEs can established. B rosilt$, Reduction Formulas for the Sine and Cosine Case L. Case2. sffidrre t g) : t 0): * sin 0 *cos 0 ,rrff-{:tcosg coswt{:*sing the To determine the correct sign on the right, it suffices to determine what mtirrnmr rI sin(ruz cos(nn cusfrr sign is when 0 is aqute. rs These are called reduction formulas since in elch case- the given expression right the on function 1 the in Case that Note reduced to a simpler expression. ii1i" ,,o*" as that o1 ih. left, whereas in Case 2 the function on the right is the cofunction of the one on the left' gthers as exercises' We wilt pgove the first of these formulas and leave the when n is even' 1X sinpe, Secondly, cos first that iin nn :0. We note nn:'(- nz is coterminal with 0 so that the cosine is * 1 and, when n is odd, nz is the coterminal with z so that the cosine is - 1. Using the addition formula for sine, we have sin(run l- + 0) Srfiuffi : - sin nIT cos 0 + cos nn sin 0 (0) cos 0 + (- 1)' sin 0 -(-1)'sin0 is This shows that sin(nn + 0) : * sin 0. Furthermore, the sign on the right g. when sign the determine ijil.*ti"t, is independent of the size of ofSo0' if we 0 is acute, EXAMPLE 11 Solution it will be correct for all values Express each Qf the following in terms of sin 0 or cos (, .| : t sin 0 (Case sin(3ru - 0) L. sin(3ru - 0) /n ror L. If 0 is acr1te,3n b. 0. c. ,ir(-;*, 1) - 0 is in *1ff:;:1;jT - 0) is positive' rhus, 332 8/ANALYTICAL TRIGoNoMETRY b. .",(; * ,) : tsin o (case 2) For 0 acute, q2 1- jis in quadrant II, and tt,. *,(; * r) : ' c. '* (-1* ,) : *cos g -sin .o.i,. is negative there. So o (case 2) For0acute,_nl2*0isinquadrantlV,wherethesineisnegative'So '*(i*r): EXERCISE SET A -cosg 4 In Problems 1-6 flnd sin 20 and cos 20' 1. sin 0 : ! and 0 terminates in quadrant II 2, cos g : -t *d 0 terminates in quadrant III 4' sec0:^'fandcsc0<0 3. tan|:Sandsin0<0 6' cscg:{"noi=e=3} s. cotg: -2ald0<0<n 7.Ilsin0:xald_n1230<nlL,findsin20andcos20intermsofx. 8. If cos x and 0 < 0 < z, flnd sin 20 and cos 20 in terms of x' 0: In Problems g-12frnd sin 0 and cos 0' g. cosz - +and ft <20 <2n 812 b. - -!^',L7r <" =+ forrnulas to flnd 13. Use the half-angle sin 1 srn20 12. tan20--+and3n<20<4n -7r<20<0 11. secz1-+and a. 10. cos c. : sin 75' d' cos 67'5' In Problems 14-17 flnd stn(ul2) and cos(a/2)' 14. cosn :*,0< a<ft a.: -*, / -I 18. If P(0) : I 16. tan 2n a u .,-3n 2\ -.), 15' sinal7' nnd P(20). lg. If 0 < 20 < 7r and P(20) - ( - *, T), flnd P(0)' sec fr -#) TtSa=+ - 3, -n 4d < 0 4IDOIJBLE-ANGLE, HALF-ANGLE, AND REDUCTION FORMULAS 333 In Problems 20-22 express the given function ln terms of sin 0 or cos 0. a. 20. frsre- sin(2n I 21. d. tth€" fu, - o) b. 0) "rr(y\2 / d. cos( -n r 0) e. ,in (-i - ,) \2 ) a. sin(g - n) b. .o, ('l - d ) $u, 22. cos(g - 3n) a. ."'(+ * e. ;. (; . , ,) b. sin(z + ,i,(-;.I d. sin e. c. /5n \ sin(r+0) \- 't ,/ .l--1; c. stn c. cos(O I ,) \- \l i g) / - 2n) ,",(, +) In Problems 23-37 prove the identities. 23. 25. 27. 29. 31. 33. r cos2x-l :SeC-r , r--2 cos"' x cos20_ sin20 srn 0 cos 0 (sinx-cosx)':1-sin2x tanz 0 II + tan'z 0: 2 r"tr 35. -t' <4m 37. B 38. sina 0 cot2 26. , I-COSZX^ -2cos2.r 29. - sina x - cos2x sin 20 -.--r^ SlN'U_T"*3tan0-tan? 2 30. 1+cos2x:sec2x --2cot20 tana * cot a:2csc2u 32. cos20 34. 2 ,or'9 21 :- cosa isin2 20 2x stnz x cot0*tan0 cot0-tan0 I 9"rr'e: 2 2(l + cos 0) -:SeC1e .0 2stn';*cos0 sin2 /. 36. sin2g 1-cosg l-tan2x .---- r l*tan'x -cos?x 0: 24. :coSo ta.o Prove the reduction formula for each of the following: a. sin(nn - 0) b. cos(nn + 0) ,,"Y#.r] e. '*lq+!r r] d. * c. cos(nn f. ,,,YTu.r] 0) s. ,,,1*+r_r] 39. If tan 0 : x and -nlT < 0 < rl2, find sin 20 and cos 20 in terms of x. 40. If sec0:xand0<0< n,0lnl2,fnd: sin20 b. cos20 c. sinj0 d. ^. writing 30 :20 + 0, find formulas for 41. By of 0. cosj0 sin 30 and cos 30 in terms of functions 334 8/ANALYTICAL TRIGoNoMETRY 42. Derive the formula sin 40:4 sin g cos 0 - 8 sin3 g cos g. 43. Derive the formula cos 40 :8 cosa 0 - 8 cos2 0 + 1,. Prove the identities in Problems 44-50. 44. cos4 3 cos 2x cos =8* z * x= 4x s 4x r-sin2 2x cos 2x 5 cos 2x * sinolr162168 1 - cos 2nx : tan nx sin 2nx 3 cos 49. sin3x*sin ff:4sin xcos2 50. By calculating sin(zi 12) in two ways, using the half-angle formulas and using the addition formulas, prove that 45. 46. - 47. x 49. lz-V3: ikl sin srn 30 0 cos 30 . - /. cos 0 sin3x+cos3x slnx*cosx - 1 -I sin 2x 6-Ol - 3- Further Identities To obtain addition formulas for the tangent, it is necess ary only to use the fact that tan 0 - sin 0/cos 0. Thus, tan(a sin(a + f) _ + B) _ cos(a + fi) sin a cos 0 + cos a sin f cos n cos p - sin ry sin B This can be improved by dividing numerator and denominator by cos a cos tan(s # fr) - cos e cos * : sinecosF*cosasinf cos r cos p cos fl cos B cosscos0_ sinasinf or tan(a f tan fi') -- B cos a cos B a*tanfi 1-tanatanfi (10n And in a similar manner, tan(e tan u-tanp - fr) = 1 + tan utan fi (11 r The double-angle formula for the tangent is obtained from (10) by takin-e : fr. If we call this common value 0, we have q, tan(0 + 0) tane+t'an0 1-tan0tan0 So, tan 20 = 2tan0 1-tanz0 5/FURTHER rDENrrrrES 335 We can derive half-angle formulas for the tangent as follows: tan Lo- |* cos |e sin Iz cos |z 2 sin 2 sin cos la ^ cos 2 Ia lx cos la 2 cosz ta Since 2 sin(a/2) cos (ut21 - cos' 2(al2) s1n - sin a and '(PI:1*cosa ** - we obtain o tan *o ;$i.: r "'-il : sln d 1+cosd If in this derivation we had multiplied numerator and denominator by 2sin(al2) 2 instead of cos(2,2), we would have obtained the equivalent formula 1 tm!a.= --cos c sln 6 (See Problem 35, Exercise Set 5.) uSe .{,, llt; L,Lr'} * We could obtain analogous formulas for the cotangent, secant, and cosecant in a similar way, but these are so seldom used that we will not clutter up our already formidable list with them. We do choose to list one more group of identities known as ths sum and product formulas for the sine and cosine. These can be obtained from the addition formulas (see Problems 32 and 33 in Exercise Set 5). Although these are used less frequently than the other identities we have considered, they are indispensable at times. Sum Formulas sinA*sin B-2sin i1 I tt-rl + siny' - sin B-2cos ryrrnry A +cos B -2 rycos T cos ,n 1L cos A-cos --''t-.*-.1 .t\.,l. .: rycos cos A+B A-B B- -Zsin2sm2 Product Formulas sin a cos p cos a sin p sin a sin B : cos d cos P - |[sin(a + p) * sin(a - il) I[sin(a + p) - sin(a - p)) +[cos(a - p) - cos(a + p)] f [cos(o( + P) + cos(a - P)1 336 S/ANALYTTcAL rRrGoNoMErRY EXERCISE SET A 5 1. Use (10) and (11) to find: 5nn b. a. tarlr, tan U 2. Find tan 20 if P(0t is the point e+,2OBl. 3. Find tan 20 if sin 0 : -i and cos 0 : -2. 4. Find tan(a + f) if sin o: t, o is in the second quadrant, cos p : -*, and B is ic the third quadrant. 5. Findtan(a - fl ifseca : 3,csca < 0, csc P : -.t6,andcosB < 0. 6. If seca:f,sina<0,csc P:+,andcosp<0,find tan(a+ fr). 7. For a and p as in Problem 6, find tan 2a, tan 28, tan !a, and tan tB. 8. Find the value of each of the following without using tables or a calculator. a.. tan b. 105" 7T tan -t2 c. tan lln U d. e. tan I95' 9. Find tan 20 and tan \0 if csc 0 : * and sec 0 < 0. 10. For the angles shown in the figures, flnd: a. tan(e + 13) b. tan 2a c. tan(a - {J) d. f. tan lB e. tan tu tan , tan 2B In Problems 11-14 evaluate by use of the sum and product formulas. 11. a. 13. 14. 15. 16. 17. t. a. a. b. a. b. sin )rft * -11 sin i- 1- 1l ll ,7nTt D. COS -1l ,3nTt D. COS - sin 105' - sin 15' sin 105' sin 15' Write as a product: COS 88 - b. b. cos 165' + cos 75' cos 165' sin 75' sin 5-t # sin 3r sin 5-t cos 3r Write as a sum: Write as a product: cos 7.r - cos 5x Write as a sum: sin 7x sin 5x Derive the formula cot(e 18. -12 t2 If cos 0 : +fr)-cotucotfr--l cot d. * cot B 6 x and sin 0 ) 0, f,nd tan 20 and tan t0 in terms of x. In Problems 19-29 prove the identities. 19. _-2 tan 2x cotx -tanx 20. o csc0-tan 1: cot 0 Tr 6/TRTGONOMETRTC EQUATTOT\S 331 11 21. l-tan? 23. 25. 27. tan 20 : 22. 1+tan atanl-secn l+tan? 2 - sin20 ti.n-^ -0 2 2sin0+sin20sinl0 - sin -i0 :lan? 26. cos 70 * cos 50 cot20-l cot20:- 24. 2cot0 sin ,4 - sin B cosA*cos.B cos 30 * _ tan * r-l -Bl cos 0 sin 30 * sin g vvr -!ih ar/..\t - ' r 28. 29. B 30" ,,,(**g),',(;-r) \4 / '.+ 30p 2cos-sin--slnl:2: . \ {i ^ rantttn+0) b. tu"[fZn-lt: Z :+tan0 I :-s.1 t rd_.1 where the sign on the right-hand side can be determined by considenng Prove the identity sin 0 sin 2e + sin 30 * cos 0 32. 1.I - Pror-e that a. 31. 2 sin 0 * cos 20 + cos 30 - H tcr be acute, tan 20 Use the addition formulas for the sine and cosine to derive the product formulas as follows: a. Add the formulas for sin(d + P) and sin(e - P) to get the formula for sin a cos B. b. Subtract the formula for sin(a - fr) from that for sin(d + P) to get the formula c. d. for cos a sin B. Add the formulas for cos(a + fi) and cos(a - P) to get the formula for cos n cos B. Subtract the formula for cos(a + fr) from that for cos(e - P) to get the formula for sin e sin p. 33. In the addition formulas for the sine and cosine make the following substitutions: a + P - A and u * B - B. Solve these two equations simultaneously for u and fr in termd of A and B, and obtain the sum formulas. 34. Prove the identity ,h sin(r * lt1 - sin r: sln --1/ taorl[., It\ .;) 2 35. Derive the formula a L-cosa tan-- _-. 2 6 sln d Trigonometric Equations The solutions of most trigonometric equations cannot be obtained exactly, and we have to settle for approximate solutions obtained through some numerical procedure, usually with the aid of a calculator. There are, however, enough such equations for which elementary techniques will yield exact answers that soine time devbted to these techniques is justified. 338 8/ANALYTICAL TRIGoNoMETRY In order to make clear the sorts of solutions we are looking for, let us consider the very simple equation 2sin0 - 1 :0 The problem is to discover all real numbers 0 for which this equation is true. We write the equation in the equivalent form sin 0 and then rely on our knowledge satisfied if, aqd only if, :; of special angles to 0 conclude that this is Tc 5n - 6, 6 or any other number obtained from these by adding integral multiples of 2r to each. To see this, remember that the sine of a number 0 is the y coordinate of the point on the unit circle that is g units along the arc from (1, 0). Thus. *. u." seeking those numbers 0 for which the y coordinate of P(0) : !. There are only two points on the unit circle with y coordinate !, and our knowledge of the 30"-60' right triangle tells us that these points ate P(n 6t and pl5nl6); see Figure 12. Since P@): P(0 +2nn), we conclude that all solutions of the equation are given by (nl6) + 2nn ot (5n16) + 2nn. Figure 12 "(?) : (+ ;) EXA\[F it is sufficient to give only the primary solutions, that is, those lying between 0 and 2n. We would know then that all other solutions are obtainable from these by adding multiples of 2zr. We can usually condense the above reasoning as follows: Usually, 1. 2. Determine the appropriate reference angle. By the sign of the function, locate all primary angles having this reference 3. Write the answer as the radian measure of these angles. angle. In our example, since we know that sin 30" :!,the reference angle is 30'. The sine is positive in quadrants I and II, so the angles are 30'and 150". In radians these are nl6 and 5n16. It might be useful to review at this time the values or the functions of 30=45', and 60'. 6,TRIGO\O\IETRIC le[ r- emr* l.l,5 tkom N sin 30" - cos ,9" : .77 Slfl-: 6 7T cos "m-{.}c tan 30o : tan 1 , tl3 6 2 7T --- 6 45' - sin cos 45" : cos sin T tan 45" - tan lt 7T1 EQrn- .\]tO\S 339 \- n arr 4lz n1 4lz It 1 4 - sim, -: cos 60' : ccs -:- tan 60' : tem '' sin 60' <! -!!\ +l 11. " ,lt <ii <tl -\ i a Also, t'ha[ rPr;s ]$ c hipk" lm n morCrr,am I A1-e Sln cos0o=cos0-1 cos 90' : cos tan 90' ft - tanl : 0 TC 2 TT 2 rs not defined less frequently and can always be obtained by the reciprocal relations.) If you feel comfortable with radian measure by now, you can omit the degree measure entirely. If an equation can be worked around to a form qrch as sin 0 : * a, cos 0 : * b, or tan 0 = t c, for instance, where a and b are any of the_numbers 0, 1, i, tl@, or $12, and c is any of the numbers O,l,11u5, or.f, then we can obtain all solutions in the manner outlined abovE. For other values of a, b, and c, a calculator or tables can be used. ;p v ]lil rhat - (The cotangent, secapt, and cosecant are not listed because they occur - # sin 90" tan 0o : tan 0 [1" 0n" l['hus+ Pq6'r amd orur $, f sin0o-sin0-0 e"ll[ L EXAMPI E 12 Findallvaluesqf 0,forwhich0 <0 <2n,satisfying 2cos2 0 - cos0 - 1:0. Solution We treat this first as a quadratic equation in cos 0 and factor: (2cos0+l)(cos0-t;:g This is true if and only if 2 cos 0 + | :0 or cos 0 : l, $ is" fiutioms &rc lr referemcs thatis, if and only if cos0:-1 or cos0:1 thome : Since cos 60' *, the reference angle in the first of these is 60". Since the cosine is negative in quadraqts II and III, it follows that we are seeking the angles 120" and 240'.In radians these are 2nl3 and 4n13. The only primaq 0o, or 0 radians. Thus, the primary solution angle for which cos 0 = 1 is set for the equation is {A,2n13, 4nl3}. 0: qgle I"s 30r'. nd 150' fln Remark. In many equations involving trigonometric functions the variable must be treated as a real number, which is the reason for writing the above answers as the radian measures of the angles. Recall that the values of sin 0, of cos 0, and the other trigonometric functions are unchanged whether we consider 0 a real number or the radian measure of an angle. An example of a mrs $-lt'. 340 8/ANALYTICAL TRIGoNoMETRY mixed algebraic and trigonometric equation will perhaps help to make this , point clearer. The equation 2sinx-x:0 is clearly satisfied when x :0, but by trial and error, using a calculator (or by some more sophisticated numerical procedure), we can find that another solution is x = 1.8955 of an angle, then the degree measure of the angle is approximately 108.6'. Now it would not b,e true thar x : 108.6'; in fact, it would not even make sense, for we would have If we think of this x as the radian measure 2 sin 108.6' - 108.6' : 0 which is a totally meaningldbs statement. EXAMPLE 13 f Find all primary solutions of sin 2x x:0. cos Solution This time we will omit most discussion and go only through the would be expected in doing the problem. sin2x*cosx:0 2sinxcosx+cosx:0 cos x(2 sin COSX:0 -L 1._ x * 1) : 2sin x- 7T 3n )' EXAMPLE 14 l* 2, 71 6.3n 2, Find all primary solutions of tan d" -1 sinx- -12 2 v- The solution set is 0 7n lln 6' llni6't. - 3 cot a:0. tan d,-3cotd-0 Solution , The reference angle is 60', tafld"- 3 r^"r-o tanza-3:0 tan d,- t^rf or 7Tl3 radians. Therefore, 2n 4n 5n N-1, T, T,T 6 steps that fI 6/TRrGoNoMErRrcEeuATroNS 341 [o make Since we multiplied by an unknown expression, namely, tan a, it is essential that we check to see that this was not zero for the values of s found, which is clearlythe case here, since tan 0. So the solution set is {z/3, 2n13, 4n13, 5nl3\. tlus a: t"l1+ catrculator (or uhat another EXAMPLE r degree meabe true thar l5 Find all primary solutions of Solution The fact that cos n :2 + sin x. x : | - sin2 x suggests that if we square both it will be easier to work with. cos2 given equation, I have sides of the 3cos2x:4+4sinx+sin2x : 4 + 4 sin x + sin2 x 4sin2 x -14 sin -r + 1 :0 (2 sin ..r * 112 : g sinx: -* 7n lln ^- 6' 6 3(1 the steps .f thaT - sin2 -t) We again must check our answers, because squaring does not necessarily lead to an equivalent equation; it may introduce extraneous roots. So we check in the original equation. When x :7n16, we get r(-*)12+(-il 33 -r=, So 7nl6 is not a solution. When x - llnl6, *(+) r 2+(-1) 33 ,:, So EXAMPLE 16 Solution x - 1lnl6 is the only primary solution. Find all primarisolutions of 2 sin 30 - .,6 tan 30 : 2sin30- .f sin 39 cos 30 0. -0 - .fl - 0 (We multiplied by cos 30.) sin39-0 | .or 30:+ (So we did not multiply by zero.) sin 3g(2 cos 30 342 8/ANALYTTcAL TRTGoNoMETRY Now we want all values of 0 lying between 0 and 2n. We must therefore find all values of 30lying between 0 and 6n. In general, if n0 is involved, in order to find all values of 0 between 0 and 2n, we find all values of n0 between 0 and 2nn, and then divide by n. We have from sin 30 : 0 that 30 : 0, TE) 2n, 3n, 4n, 5n and 0:0, ;,+, From cos 30 : ,ll Tc, +, + lZ, we have lln 13n 23n 25n 35n 30: rE 6' 6' 6' 6' 6' 6 TE lln l3n 23n 25n 35n 0- 1g' 18' 1g' 1g' 1g' 1g In both cases we found the first two values then 2n again. The complete solution set is 7E {0, EXAMPLE 17 #, 5' of 30, then added 2n once, and lln 2n L3nn' 23n 4n 25n 5n 352) 1g 'T' 18 ' 1g 'T' 1g 'T' 18 -\ J Find all real solutions to tan x - cot x:2. tanx-cotx:2 sinx_cosx_,) Solution COSX SlnX sin2 x- cos2 2x:+, 3n g) x- 2sinxcosx - cos 2x - sin 2x tan 2x: -1 7n lln L5n (We "went around" T, 4 , 4 7n lln 1,5n T' 8' twice.) 8 As none of these values of x makes sin x, cos x, or cos 2x eqtoal to zero, we did not multiply by zero. We must also check to see that we did not lose any roots when dividing by cos 2x;that is, we must check to see if any of the roots of cos 2x: 0 are roots of the original equation. Since cos 2x : 0 when 2x : nl2, 3n12, 5n12, 7nf2, and, so when x : nl4, 3n14, 5n14, 7nf 4, we check and flnd that none of these is a root of the original equation. So we have found all primary solutions. To get a// solutions, we add arbitrary multiples EXERCI 6rrRrcoNoMETRrc EeuArroNS 343 herefore flnd YEd. d of 2n to each of these. So the complete solution set is in order 7n l,ln l5n l9n 23n 27n 3ln 35n 39n 43n 47n !v ' T' g' g' g' g' g' g' g' g' g' g' between 0 ls II ) or we could write \3n18 * 2nn,7nl8 * Znn, llnl9 * 2nn, l5nl8 * 2nn: n:0, +1, +1. We could continue with examples, each possessing its own special features, but these illustrate the main techniques. It might be useful to summarize some of the points that the examples were meant to bring out: 1. h'once, and 2. 'I x: *b, and so on. Find the radian measure of all primary angles (that is, 0 < x < 2n) satisfying the elementary equations in 1. If all solutions are desired, add arbitrary multiples of 2n to each of these to obtain the complete solution set. 4. EXERCISE SET A to zero. wE m( trose an!' o[ 0$rE !; :0 wfo66 h me chesk io cos 3. If in arriving at the elementary equations both sides of an equation were J anr- Use algebraic techniques such as factoring, multiplying by the LCD, and squaring in conjunction with basic trigonometric identities to simplify so as to obtain one or more elementary equations of the form sin x : * a, rr.e liar e ry muluiptres squared, or if both sides were multiplied by an expression containing a variable, it is necessary to check the answers. (In the case of multiplying by an unknown, it is sufficient to check that the multiplier was not zero.) If both sides of an equation are divided by an expression containing a variable, then the values of the variable for which this expression equals zero must be checked in the original equation to see if they are solutions. (In general, such division can be avoided by factoring.) If the final elementary equation(s) is of the form sin nx : * a, cos ,4-r : +b, and so on, then to get all primary values of x, find all values of n-r between 0 and 2rut, and divide by n. This amounts to finding the angles in the first revolution and then adding2n a total of n - 1 separate times and finally dividing everything by n. 6 Find all primary solutions unless othenvise specifled. L. 2cos x-J3 3. secx -2 5. cosx-2coszx:0 7. cos2x*cosx:0 9. sin2/-cos2t-l 11. 2cos x- cotx-0 13. 4tan2a:3sec2e 15. cos2 x- sinx*5:0 2, cotx*L:0 4. sin2 x - 1 (Give all solutions.) 6. sin20-cos0:0 8. 2tanrlr-tan$secr/-0 10. sec20:2tan0 12. 2cosz x+ sinx- 1:0' 14. sin2 x- 2(cosx- 1) 16. tan 2x - cos 2x * sec 2x - 0 r 344 8/ANALYTICAL TRIGONOMETRY x : 17. 1, -.cos smx 19. sin 60 * 21. tan!* 2 sin x sin 30 cos x : : 0 B 26. sin g * cos 0 : cos 20. sin2 4e - cos 20 sin 40 +1 -0 - 2 -0 2 : 24. O 2cos x- cotxcscx -0 | (Give all solutions.) sin x :- cos x 28. 1-cosx 1+sinx 30. 13 tan 0 :2 sec g - \F cos 2x - 0 29. sin 20 - 4sin 0 : 3(2- cos 0) 27. 2 cosz 20 - 3 22. sin2!-sin2x:0 I 23. 2(sinz 20 - cos2 20) + f 25. 2tan2g+sec0+2-0 18. 2x tan 3x + 1 In Problems 3I-36 use a calculator or tables to find all primary solutions correct to three significant figures. 32. cos2x+2cosx-0 Y. 3sinx-tanlx-0 36. secx-2-L0cosx 31. 12sinzx-sinx-6-0 33. 2sec2x:5tanx 35. 2sinx-cscxf 3:0 In Problems 31 - 3g find all primary solutions without using a calculator or tables. 38. 37. tan3 x*tan2 x- 3tann-3:0 39. 2sin3x*3sin2n-1:0 7 Graphs of the Trigonometric Functions To obtain the graphs of y : sin x and.), 8sina 0-Zsin2 e-3 -0 : cos x we make use of the definitions of the sine and cosine as the ordinate and abscissa, respectively, of a point on the unit circle (Figure 13). Notice that we have used r to represent arc length on the circle from (1,0) to P(x) rather than the customary 0. We have used capital letters to label the axes. Table 1 indicates how sin x and cos x vary as x varies from 0 to 22. These facts can be verified by visualizing the point P(xl moving around the circle in a counterclockwise direction, starting at (1,0). Furthermore, it should be clear from considering Figure 13 that the ordinate Figure 13 P(x) (cos x, sin 7/GRAPHS OF THE TRTGONOMETRTC FUNCTTONS Table 345 I As x goes from A to nl2 nl2 to n n to 3ny2 3nl2 to 2rc sin x goes from 0to 1to0 0to -1 1 -1to0 cos x goes from 1to0 0to -1 -1to0 0to 1 and abscissa of P(.r) (that is. sin -x and cos .r) \'ar)- in a uniform way, with no breaks or sudden changes. as P(-rl mo\-es around the circle. This information. together with our knosledge of sin.r and cos x for x : nf 6, nf 4, rl3 and related values in the other quadrants. enables us to draw the graphs with reasonable accuracy. Since sin -r and cos -r each has period 22, the graphs repeat every 2n units. So once we know the graphs in the interval from 0 to 2n, we can extend them indefinitely in either direction. The graphs are shown in Figures 14 and 15. Dfus Figure 14 UilTMS ist Figure 15 om @iltu ruM Ey &$ ,fro,,, ;[, (illttr" lire The maximum height attained by the sine curve and by the cosine curve is calledtheamplitude. So, fory: sin xandy: cgs xtheamplitudeis 1. To obtain the graph of y : tan x we again refer to Figure 13 and use the fact that tan x : sin x/cos x. Therefore, we consider the ratio of the ordinate of P(x) to its abscissa as P(x) moves around the unit circle. When x : 0, tan x : 0ll: 0, and as x increases toward nl2, the ratio of the ordinate of P(x) to its abscissa steadily increases, taking on the value 1 at x : n 14. For x near n 12 the ratio becomes very large and gets larger and larger without limit as x approaches.nl2. At x: nl2 the ratio is not defined. For x slightly greater than nl2 the ratio is negative, since the ordinate is positive and the abscissa is negative, but its absolute value is large. At x:3n14 the ratio 346 8/ANALYTICAL TRIGONOMETRY is - I, and at n, it is back to 0. Since tan(x * n) - tanx*tann 1-tanxtann Fis tanx*0 1-(tanx) '0 =tanx it follows that tan x has period n. So we can draw its graph between 0 and n and duplicate this in each succeeding interval of length n. Similarly, we can extend it to the left. With this information and our knowledge of tan x for special values of x such as z/6 and nf3,we can draw the graph, as in Figure 16. The vertical lines at odd multiples of nl2 are asymptotes. Figure 16 ris Figure 17 )/-cotx The grapfu of y Figure 17. : cot x can be similarly obtained and we show ils graph in ofx is defined as the reciprocal ofthe xcoordinate ofP(x), that of cos x. Since lcos rl < 1,it follows that lsec xl > t. ttre reciprocal is, as the x are the same. When cos x : l,sec.x : 1, and when x cos of sec and signs : This latter occurs at x : nl2,3nf2, -n12, 5nl2x is undefined. .x 0, sec cos gets cos x gets close to 0; so sec x gets arbitrarily nf2, to x close on. As and so with the values of sec x for x : nl6, nl4, nl3. together analysis, this large. Using Figure 18. The graph of7: csc x can be graph in the on, we sketch so and similarly obtained (Figure l9). Both sec x and csc x have a fundamental period of 22. No amplitude is defined for them. They each have asymptotes; for the secant they are the lines x:n12, -n12,3n12, and so on, and for the cosecant the lines x:0, ft, -fr, the secant 2n, and so on. 7/GRAPHS OF THE TRIGO\O\IETRilC F-{--\CTIC\S il7 Figure 18 :gnmemC rly. U IV r" 2n \,n-e .,eu. - 5nl2 lf, tam "r icr I Figr,lne 16 - 3nl2 - 7T12 3n12 5ru1 Figure 19 Figure 20 AmplituJ. -, Tc+-x n12 f gne,pr P(xt" Iad rirl E4 Period -;l Amplitude - 1 i Period---l f Period-> rm 1 Period ::irrrii: w;'klu' _\il a rtiw"--u ,a[,{r3.r cfllr :(8 dinme r$ what we have shown are the basic graphs of the trigonometric functions. You should become especially familiar"with the fust three. The crucial facts about the sine curve are illustrated in Figure 20. This, together with the knowl_ edge ofthe periodicity, enables us to obtain a rapid skJtcn of the curve. For the basic sine curve /: sin x, the fundamentar period is 2n and the amplitude is 1. Now we want to consider the effect of introducing positive constants aand.b: the hs$ I m- -n- Y- astnbx we will take one at a time. First consid er y q sin x. This has the effect of multiplying every y value of the basic curve by o. So when the basic curve is at its maximum height of 1, the new curve will be a units high. Thus, the ampli- 348 8/ANALYTICAL TRIGONOMETRY Figure 21 ! : Amplitude A sln x + tude becomes a, and the period remains unchanged, as shown in Figure 21" We have exhibited the curve for one period only (also called one cyclQ, since the extension is obvious. Next, let us consider./ : sin 6x. We know that the sine curve completes one 22. But then n goes from 0 to 2nlb. So,we conclude to from Znlb. This can also be seen in another way. To sa1-. period of sin 6x is that the that sin x has period 2n means that sin(x + 2n) : sin x. Thus, sin(6x * 2n) : sin 6x. But sin(bx 1- 2n) : sin 6[x + (2nlb)f. So, cycle when 6x goes 0 / sin a(x r-\ .+): sin 6x Thus, when we add 2nlb to any x, we get the same value as sin 6x. That is. sin bx has period 2nlb.The effect of the multiplier b in this position, then. is to alter the period; it is shortened if b > I and lengthened if b < I (Figure 22t. At' t' Figure 22 -)' .\mplitude I - 1 : sin bx 2nlb nlb 7rl2b Remark. When the variable x represents time, the period is the time required to complete one cycle. The reciprocal of this, called the frequency, gives the number of cycles (or fraction of a cycle) completed per unit of time. Thus, Frequency: rb : Period t This is a widely used concept in electronics. Now we can put these two changes together to get the graph of y : a sin D-t. as shown in Figure 23.If , in addition, a constant k is added to the right-hand side, giving ! : a sin bx * k, this has the effect of shifting the entire graph of ! : a sinbx vertically by k units-upward if k > 0 and downward if k < 0. We consider next an equation of the form y-asinb(x-a) (11 n 7/GRAPHS OF THE TRIGO\O\IETRIC FU\CTIO\S y9 Figure 23 Amplitude: a stn bx a I v f Period ilr,"0---l n Figure Period 1 Period a * b(* ome - - 2nlb----> a) equal to 0 gives b(* - a) x* a- Tia)-. To say - - I The sine will complete one cycle when b(* we conclude (Dr Amplitude 11. crctrg, simm mpletes ] arfl : - a) goes from 0 to 2n. Setting 0 0 X-(X and setting b(* - a) equal to 2n gives b(*-a)-2n bx- That us" [bn, them. is (Figure -l I x- O: b x: 2n +d' b So a complete cycle occurs in the interval from ato (2nlb) * a. This is a distance of 2nlb, so the period of 2nlb remains unchanged, but the curve is shifted a units horizontally. If a > 0, the shift is to the right, and if a < 0, it is to the left. we call a the phase shift, and say that the curve is lal units out of phase with the curve / : a sin Dx. This is illustrated in Figure 24. Finally, we note the effect on the graph of ! : a sin Dx if either a or b is negative. If a < 0, every y value is the negative of what it would have been if a were positive. For example, in y : -2 sin Jr, every y value is the negative of the corresponding y value in y :2 sin x. So the effect is to flip the graph of y:2 sin x about the x axis. lf b < 0, we make use of sin(-0): -sin 0. irc requeuud ftr glvm thr m- 2n Thu"'. Figure 24 !: a sin b(x - (2nlb) f-dsilrlSru c right-h*qr*dr lirc graph str rd if& < # 1 r Xltn a) Period - 2nlb +a 350 8/ANALYTICAL TRIGoNoMETRY For example, we would write y: sin(-2x) /: as -sin 2x, and proceed as above. A srmilar analysis to that for the sine holds for the cosine function (se Figure 25). Note, however, for the cosine that cos(-0): cos 0, and so the graph of y : cos(-2x) would be identical to the graph of ./ : cos 2x. with appropriate modiflcations because of the different period (for the tangent and cotangent) and lack of amplitude, the other functions, too, could be analyzad in an analogous way. You will be asked to consider certain of these in Exercise Set 7. Figure 25 !:acosb(x-o() Ampli (Znlb) + u x It rfl 1 Period shift - E,XAMPLE 18 Sketch the graph of y :3 a sin - fx. Solution The amplitude is 3 and the period rs 2n + Figure 26 EXAMPLE 19 + Sketch the graph of y - 2nlb + - 4TE (Figure 26). l 2 sin(3x - TE). Solution We first factor out the 3 to put this in the form of equation (12): /n\ sin'f -, "r:2 This is therefore a sine curve with amplitude 2, period 2nf3,and phase shifr z/3 units to the right (Figure 27). n"t qu 7/GRAPHS OF' THE TRIGONOMETRIC FUNCTIONS d proceed as fuilction 3s1 Figure 27 (see t, and so the cos ?.r" Wiffi e tangent and il be anallzed se in Exersise EXAMPLE 20 Solution Sketch the graph of y- cos12* + (nl3)f. We first write the equation in a form analogous to equation (12): ./:cos So the phase shift u The amplitude is l, rQ*I):"or r?-( .J is - nl6. The curve is therefore shifted zr/6 units to the left. and the period is 2nl2: z (Figure 28). Figure 28 2t: EXAMPLE 2I Solution rnd phase shift Sketch ! :2tan$x. There is no amplitude defined for the tangent curve, but the effect of the coefficient 2 is to multiply all y values of the basic curve by 2; in particnlar, when the basic curve is I unit high, the new curve will be 2 units high. Since the fundamental period of the basic tangent curve is z, the period of this curve is z + | :2n. We show two complete cycles in Figure 29. 352 8/ANALYTICAL TRIGoNoMETRY Figure 29 8 EXERCISE SET The Invet 7 In Problems l-25 sketch one cycle of each of the curves. Give the period and, where appropriate, the amplitude and phase shift. A L. y-stn2x 3. y-2sinx 5. ! :3 cos(zrxl2) 7. !:2sin(-3x) 9. y - -sin(xl2) 11. y- sin2x* 13. y-2sin3x-l 15. !:ttannx ) !: cos 3x y -2stn3x Z. 4. 6. * 8. 10. ! : -2 cos x 12. !:2 cos x - 3 14. y-3cosnx*2 16. y - 2 cot(xl2) 1 B : / z\ -;) 17. !:rrr(r-;) 19. y 19. !:sin 2(,.;) 20. !:cos 3(,.;) 22. y- 24. / !:!tan(r, 21. y-2sin(3x+2) 23. !:1,-2rrn(\ n*-+) 4/ 2s. y_ 2+2.or(r, 26' a' Figpe y - 4 sin(nxl3) y :3 cos( -TEx) cos(x 3cos(3 -2x) z\ -;) ) il,t*i'f'ill3""1.f dividing by ",trTTE GTtr show that a sin x sin(x + 0) where 0 is as shown in the accompanying sketch. * b cos x can be 8/THE INVERSE TRIGONOMETRIC FUNCTIONS 353 b. Use the technique of part a to !: What is the value of 0 in this -*J 6,.1-r,lsl \\ ,lIC. "--L--- of sin x + \rc cos x case ? 27. By the technique of Problem 26,parta, write each of the following as a sine function. Determine the amplitude, period, and phase shift. 29. Discuss the effect on the basic secant curve of introducing positive constan ts a and b to obtain y - a sec bx. 29. Sketchy-2sec(xl3). L. y-sinx-cosx 8 analyze and sketch the graph b. !:3cosx-4sinx The Inverse Trigonometric Functions Let us look again al the graph of .r' : sin .t (Figure 30). It is immediately evident that this is not a one-to-one function. so it has no inverse. However, by a suitable restriction on the domain of the sine. an inverse can be found. The standardchoiceistorestrict,.r-sothat -r 2 < x < r 2. Thesinecurvewiththis domain will be called the principal part of the sine cun,e. = Figure 30 L3nl6 Ifx is so restricted, then for each y such that | < y < r, there is exactly one r such that y: sin x. For exampie, if we take-y : L, we get x : nl6. On ttre other hand, if y: -t, then x : -i16. The equation :iin x expresses y in "y terms of x; we would like to solve this equation for x in terms oly. Unfortunately, we as yet have no way of doing this. with the aid of the graph (or tables or a calculator), we can find x for any given y, but we have no simpte equation that expresses x in terms of y. what we do is invent a symbolism, ;:: - x=sin-1"/ Jrr which is real,o'x is the inverse sine ofy." Actually, we should probably read it "principal inverse sine of y," since it is the principal part of the sine curve that is used in finding x. Unless otherwise stated, we will in the future understand that sin- 1 y means the principal value. as the There is another symbol in wide use that means the same thing as itis sin-1y; arcsinyrread"arcsineof y." Itsoriginprobablyliesinthelengthof arcon a unit circle used in the definition of the trigonometric functions. Thus, arcsin I might be interpreted as "the length of arc on the unit circle for which the sine 354 8/ANALYTICAL TRTGoNoMETRY ir 1." We know the length in this case is zr/6, so .TE sm6:, 1 7T and arcsln 6 1 , are two ways of viewing the same fact, The first says that the sine of the number nl6 is t. The secortd says that nl6 is the number whose sine is |. In what follows we will use both notations for the inverse sine (as well an ur inverses of the other trigonometric functions), since both are in wide use. consider some examples. kt EXAMPLE 22 Solution Find the value of: a. sin-1 a. sin- 1 11 : b. sin-'(-r) c. sin-10 nl2, since nl2 is that value of x on the principal part of the curve whose sine is sir 1 b. sin-11-j1 : -116 It is important here to note that the answer is not 112/6, though an an_el,. of llr,6 radians and an angle of - nl6 radians are coterminal. The distinction is most clearly seen by looking at the graph of the sine curve. On tbc .r axis. - tt16 certainly is not the same point as llnf 6, and even though sine of each is - l, we choose - nl6 since it falls within the restricted rangr on -r that defines the principal part of the sine curve. tb sin- 1 0 EXAMPLE 23 0 Evaluate: a. arcsin J3 2 Solution - a. arcsin th :n 23 b. arcsin(#) b. arcsin (#) : -4 - c. arcsin( c. arcsin(-1) 7T 1) : -: 2 To plot the graph of the inverse of the principal part of the sine curve, $e follow our usual procedure with inverses of solving for x and then interchanging x and y, giving -Y : sin-1 x Now, since the roles of x and y have been switched, we must restrict x so that -1, <x < 1 andy will fall in the range -nl2 <y <n12. The graphis just the reflection of the principalpart of the sine curve in a 45' line through the origin (Figure 31). Any value of x between - 1 and 1 corresponds to a unique value 8/THE TNVERSE TRTGONOMETRTC FUNCTTONS 355 Figure 31 n - Utl' rrmLttr -1 Tcrc \ I rl2 ; Urrel .tS s- Ler ;s 1 _nl4 I _1 1l (x( \-"lz SyS "l2l - 7T12 &E of y between -nl2 and nl2.When x is positive, y is positivg and when x is negative, y is negative. Notice that y: sin-1x if and only if sihy: x and 1 -nl2 < y < nl2. So ryhen we wish to evaluate sin- x for a particular x, we ask what number (or what angle in radians) between -nl2 and nl2has x as its sine. We consider next the inverse of the tangent function, since it has much in common with the sine. The principal part (or principal branch) of the graph of ! : tan x is that portion between -nl2 and nl2 (Figtre 32). On this portion, $,yr#r m m$Iffi : dimrumu- if any value ofy is specified, a unique value of x is determined. Solving y for x leads to either -&ffi.tc ryfumr g:tan-r y ud rmrryrr or : tan x x - arctan y Figure 32 ,m .rr Again, we interchange the roles of x and )), and write m"mu !: tan-1 Y or |: atctan x hrrouu*ilqm and restricty such that -nl2 < y < nl2. The graph of this function is shown in Figure 33. With our knowledge of special angles, we conclude, for example, that "milfuwl $m dfihn: E@mr rc mdihro tan -1 , -fi 4 tan- 1(- 1) : -! 4 _1 tan I G7T VJ :; J tan-t g - 0 356 siANALYTICAL TRIGoNoMETRY Figure 33 A consideration of the graph of the cosine function will make it clear thar some different portion will have to be used as the principal part (Figure 341 because when x lies between -nl2 and nl2, y is always positive and hence do* not assume all of its possible values. Furthermore, for a giveny there is generalll' not a unique x determined. We choose instead the portion of the curve lying between 0 and a as the principal part. Then, for eachy such that 1 < y S 1there is a unique x for which y cos x. Again, we express the dependence of -r on y by writing - : x-cos'y _1 x - arccos y or 1 Interchanging the roles of x and y, we obtain the graph of y : cos - x (Figurt 35). When x is positive, y is positive and between 0 and nl2; when x is negatir-e. y is positive and between nf 2 and n. Figure 34 v Figure 35 ft [-t(x( I o<v< t) 2nl3 n12 _l I:cos'x n14 - 1- Llz u,l1 The inverse cotangent is seldom used, but is defined in much the same way in Figure 36. The secant and cosecant present certain problems, and there is no general agreement on what parts of the curves to invert, that is, which portions to define as the inverse cosine. Its graph is shown 8/THE TNVERSE TRTGONOMETRTC FUNCTTONS 357 Figure 36 !-cot --r i lhemce Coes is gemeraltr3 fl- ndeunce ofr'r I r' ilFigure ir negaive- -;;.;r o<v<n I : crrrr-e hrmg -I < l'< ^x as the principal branches. Fortunately, this lack of agreement is not serious. The inverse cosecant is seldom used, and since it can always be circumvented. ue will omit it entirely. The inverse secant is sufficiently useful to warrant consideration. For purposes of later use in calculus, there is some justification for the choice of principal parts made below. Let us consider first the graph ofy sec -r (Figure 37). No single branch of the graph is suitable for defining the inverse function. While the reason is not now apparent, we select the highlighted portions in Figure 37 to define the inverse function. Thus, when we take 0 <x < nf2,and wheny < -1, we take TE<x <3nl2.When we interchange the roles of x and y, we obtain the graph shown in Figure 38. ft cilear ruhar (Figure 1S*r _1 y2l, - Figure 37 - 3nl2 - nlz irlz TE 3nl2 A lAl Figure 38 ,' 7q 11 \. -l::'- 0<y <nl2 [*rr, [x< -l,n<y<3n12) --TE12 -1 EXERCISE SET Etr A EBmemaI Btodefinc 8 In Problems l-32 give the Fme wlar 2n -1 1. a. sin- 1(- ,fTtzl 2. a, tan- '(- 1) 3. r. arccot( - 1) value. b. b. b. arccos(-V3i2) arcsec( sin-1 1 -2) ] 358 8/ANALYTICAL TRIGoNoMETRY 4. 5. 6. 7, 8. 9. a. arccos I b. tan-1 I a. arcsec I b. sin-1(-) a. arctan0 b. cos-11-1; r a. arccos0 b. sec-L(-21"f3) u. cos d lf a: sin-1(-f) b. tar0 if 0 : arccos(-lo) a. sin x if x : sec-1(-3; b. sec 0 if 0: tan-1 f; 10. a. sin[cos-1(-$)] b. sec[sin-1(-1n)] b. cos[tan-11-$] 11. tan(arccos!) 12. ^. a. sin(arctan fi) b. csc[arccos(-zJ] 13. cos[sin-1(-])l b. cot[cos-1(-fr)] 14. ^. sin(a - p) if a : sin-r t and P: cos-1(-:13) 15. cos(a + B) if a : arctan(-l) afi P : sec-1 i 16. tan(u+ $ if a: sin-11-$; and p :cos-1(-p1) 17. sin20 if 0 : cos-l(-t) 18. cos 20 if 0 : sin-1(-+) 19. tan20 if 0 : sin-'(--5-rJ 20. cos[2 cos-l(-!)] 21. tan(2 tan- t t) 22. sin[2 sin - 1( - f)] 1(-f)] 23. cos[2 sin25. cos[] arcsin( - f)l n. sin g and cos 0 if 20 24. sin$ cos - 1 fi) 26. tan +[tan - 1( - )] 125 : tan-r(-at) 28. sin(al2), cos(al2), and tan(al2) if a : 29. sin(sin-1 {f + cos-'!; 31. tan(tan-1 l+tan-L 11 1 cos - 3 30. cos[cos-l S - sin-l1*fi;] B 32. tan-t l - tan-1(-$ Hint. Call this u - B and find tan(d - 0. 33. If 0 : arcsin -r, show that xi.Il? : tan g. y, If 0 : arctan .r-. show that , 1 + 12 .r2 : csc 0 cot 0. 35. Show that the expression ....2 3. -I7l 9 changes to sec g - cos g when the substitution 0: tan-1(,r:3r is made. X. Show that . E -r : sin 0 under the substitution 0: sec-r(xl2). o 7 Trigonometric Form of Complex Numbers f ii*?:11?'#3,,::",:"i..Tl':T'f Br:T,i1ffi::.T#l:11i3::J:#.: with this the point (a, b) in the plane. As in Figure 39,let r denote the distance from the origin to the point (a, b), and let 0 denote the angle from the positive x axis to the line from the origin to the point (a, b). Then we have r=ffaF a=rcos0 (13t b=rsin0 The complex number z : q I bi can therefore be written in the form s-r(cos0+isin0) (14i L\ \\IT g/TRIGONOMETRIC FOR\{ OF CO}[FI-E\ \[- \IBERS 359 Figure 39 This is called the trigonometric form (or polar form) of the complex number :. The form a * bi is referred to as the rectangular form. It should be noted that the trigonometric form is not unique, because if any integral multiple of l; is added to 0, the value of z is unchanged. For example, if - 2('"' \+"t";) I -3r then also / 7n i sin /-/ ('o'T + --t) 4? 2(,", + *isin +) and so on, where integral multiples of 2n are added to the angle. Normally, we choose 0 to be between 0 and2n, but there are exceptions. The number r is called the modulus of z and is often denoted by lzl. Recall that for real number x, the absolute value lxl can be interpreted geometrically as the distance on the number line between the origin and the point representing r. Similarly, lzl represents the distance in the plane between the origin and the point representing z.Infact,lzl is sometimes referred to as the absolute value of z. The angle 0 is called the argument of z. t-xrl b T)- zubstiirutiom EXAMPLE 21 Find the trigonometric form of the complex number , : modulus and argument of z? Solution J3 f i. What are the The number \fr f i is in rectangular form a + bi,wrth a: ,13 and b r-vlr+1-z,and0- TEl6 (Figure 40). Therefore, by equation l4l, ( ryrnbers amd ,[*,rt";) - z("o \ ffe associate z the distanse the positire ' The modulus of z ts 2 and its (primary) argument is nl6. ) I Figure 40 tt3n fl["** t. So 360 8/ANA LYTICAL TRIGC}NOM ETRY One of the advantages of the trigonometric form of complex numbers is that products, quotients, powers, and roots are particularly easy to calculate with numbers in this form. Consider first the product of two such numbers' say : zr: rz(cos 02 + i sin 0r). On multiplication. : making use of the fact that i2 - I and arranging terms, we have ZtZz: rrrr[(cos 0, cos 0, - sin g, sin 0r) * i(sin 0, cos 0, + cos 0, sin 0r)] zt rr(cos 0r + i sin 0r) and By the addition formulas for the cosine and the sine, this can be written in the form zrzz= rrrr[cos(O, + e) + isin(01 + 0)f (l5l So the modulus of the product is the product of the moduli, and the argument of the product is the sum of the arguments of the two numbers. EXAMPLE 25 THEORT Find the product of the two comPlex numbers zL Solution DE MOII / ir\ - 2(.o' i+ i sin,I) and zz :: (.o, + - rt, ;) \ By equation (15), the product is z,.zz:2 3 *i sin(;.;)] [cos(:.;) : o ("o, i* t'n\) This can be put in rectangular form by evaluating the trigonometric functions: ,' l):0 * 6i:6i Ztzz:6(0 + In a similar w&y, we can prove that tf z, Z:A EXAMPLE 26 - 0r) 0, then + isin(0r - 0r)] (15 Find the quotient z, f z, rf zL- 4t", Solution [cos(g1 * +.i sin?) zr z2 and zz:2(.", i+ t ri";) ::['"'(+ -;)*isin(+ il] - z(.", \+- "";) EXA}TP r g/TRTGONOMETRTC FOR\I OF CO\[PLEX fiers lS -;ilirr hulam w*":r-T. In rectangular form this hiplraail'lrtr- ritten ::- l - is Z:rG*,+) :1+i"3 mhem,* su',& ffr s,n r, \U\{BERS 361 It is in raising to a power that the trigonometric form has the grearesr adr artage. We will state without proof the following important result due rr-r rhi French mathematician Abraham De Moivre (1667-1754): rruF :: 4ummud I MOME'S THEOREM DE Let z= r(cos0+isin0) be the trigonometric form of any complex number. Then for any natural number z 7" = r"(cos n0 + i sin n0) Thus, the modulus of zn is the nth power of the modulus of z, and the argument is n times the argument of z.That the result is plausible follows from a consideration of a few powers of z. By equation (15), $) z2 : r' r[cos(O + 0) + i sin(0 + 0)] : r'(cos 20 + i sin 20) s Similarly, applying equation (15) again, 23 ;fumcnlcrf,$ : : z2 .z: .rlcos(2O 12 + 0) + isin(20 + 0)f ,'(cos 30 + i sin 39) This process could be continued, and for any given power of z,De Moivre's theorem would be confirmed. However, this does not co4stitute a proof of the theorem. It can be proved using a technique called mqthematical induction. which we will study later in this book. t, j 6r,'i EXAMPLE 27 Expand (1 + i)u. Solution With the aid of a sketch (Figure 41) we determine that r: Writingz:l*i,wehave z6:(r + il6 *I - / = Gl46 (cos Vt"' i+i sin;l' 6n _, 6r\ q +i sin 4) _. -8t",+.i sin +) In rectangular form the answer is (1 +i)u -8[0+i(-1)]_ -8i Ji and A :; { 362 r 8/ANALYTICAL TRIGoNoMETRY Figure 41 Remark. You might wish to compare this solution with the work involved in dding this problem by expanding by the binomial theorem and simplifying the result. You will find that the method we used is much simpler. Finally, we consider the problem of taking roots of cornplex numbers. Again : r(cos 0 * i sin fl, and suppose we wish to find <li.W" first recall that if any integral multiple of 2n is added to 0, the same value of z results. That is. for any integer k, let z z : rlcos(O + 2kn) + i sin(0 + 2kn)l Making use of De Moivre's theorem we can prove the result ;1,' = rr i,[cos (ry) . i sin (l7t W) Furthermore. there are exactly z distinct roots that can be obtained by taking k = 0, l, 2o . . ., n - l.These z roots are equally spaced on a circle of radius r1''- No.:. In working with the trigonometric form of complex numbers, the angle 0 may be expressed in either degrees or radians. Ifdegrees are used, then equation (17) must be written ,, n -,,,n[.", EXAMPLE 28 Solution Find all cube roots of Let z : -8 : -8 - * i sin (,.*, (*+*) 8. + 0i. Then r : 8 and0 : r(Figure42). So byequation(17t. fi - z, z- 8ra [co, W) . i sin ej1 and the three distinct roots correspond to k :0, k : 1, and k and (, be the roots corresponding to these values offt. Then (o : 2G". )] : i * i sin N ( (t : 2f"r- n*2n * i sin :2. Let (0, (r. r(i., +) :r + ;.'/3 z*22\ ): - 2(cos 7t + i sin n) : -2 EXER.( e/TRTGONOMETRTC FORM OF COMPLEX (z:z(.o, *#+ir,"$3) -z(.", +.i NUMBERS 363 sinT) -2G +)--1 -ir3 These roots are shown graphically in Figure 43. nnohed im ffiins &c fs" Figure 42 AgEsrE nll thar ff i That ln Figure 43 1 + ir13 x * l*r$i bf (z t-rn* -1- iT frrn'*- bcrr tfuE tr6d- tt?,*ilm EXERCISE SET A 9 In Problems l-10 find the trigonometric form of the complex number. Draw a sketch in each case. 1. l-i 3. 15-i 5. -4 2, 4. 6. 8. 7. -z - ziJl 9. 5 rim I tr-L In Problems 1r. cil (m,- t'o- / z(co' t+i^15 -8i +Ji - qiJ, 10. i-l ll-20 find the rectangular form of the complex number. 4n 4z\ f + isin!) / 3n 3z\ 13. t(."r;+isinr) rs. u("orA* 6 \ ,.,n It) 6) - / 7n 17. Ur(.or?+isinZ/ ':) 19. 2+2i 3(cos 150' * I sin 150') / ln + isin,3z\ t2. *f"' ; ) 14.7(cosz+isinz) 16. / 5n 5z\ '('totT+isinr/ tB. 5(cos o 20. $(cos 315" + i sin o) * i sin 315") 364 8/ANALYTIcAL TRIGoNoMETRY In Problems 21-27 find z, . z, and zrf z, using equations (15) and (16). When possible write answers in rectangular form. a calculator, tables or without using 2t. zr r\ / n : 2(cos 22. zr : U+ S(cos z * i sin;1, i sin zz z), ,, -- / 5n+ isin ..5r\ : 3(cos O1 4 +(orT. ,t^T) /nn\/3n3n\ zz : 2(cos i+ istn 4 ) )+ i sin)). A. z, : J2 + iJ2, zz: 4',11 - 4iJ, 25. z, : ,13 - i, zz: 2 + 2i,,,t3 26. zr : 3(cos 70' + i sin 70"), zz:5(cos 50'+ i sin 50') n. zt : 4(cos 100" + i sin 100"), zz: 6(cos70' + i sin 70') 23. zr : 4[cos In Problems 28-34 find the indicated power using De Moivre's theorem. Express answers in rectangular form. [,(...|.,,r,f)]. 240)]s + 30. 28. , sin [2(cos 240" 32. 34. B [r l)'o t-O- i"Dl' [,(.,f *,.,,f;)]' 31. (J3 - 0'33. (2 - 2iJr4 In problems 35-40 flnd all the roots indicated. Express answers in rectangular form wheneter this can be done without tables or a calculator. 35. Cube roos of 8i 3j. SixthrootsorL 39. Square roots In Problems of 4l-M - l6i 45. 36. Fourth roots of ;r. -l - '€ -?.,?A 40. ";;;rootsor Fifth roots of - 16'18 + 16i frnd the complete solution set. 41. x6 + 64 -_ 0 43. x6 : 64 42. 44. xa x3 + 256 :0 + 27i:0 Prove equation (17). First wiite z : r[cos(0 + 2kn) + i sin(0 + 2kn)f, and let ( : p(cos d + ,sin d) unknown nthrooiof z. By De Moivre's theorem determine p and $ such that Hint. be an rt_- 10 zs. Polar Coordinates In all of our graphing so far we have used the rectangular coordinate system. For certain g.upt irrg problems, however, an alternative system, called the polar coordinateiystem, has definite advantages. In this system points in the lo/PoLAR cooRDIN, a TES 365 planearelocatedbygivingradialdistancesfromafixedpoint,calledthepole, extends horiirrJ ungt., from a i*ia.uV called the polar axis. The polar axis 1',r pesffi3l{g ,"tt"ffytotherightfromthepole'u'd*tmakeitintothepositivehalfof Now let 0 be any angle a number line, with the pole correspondin g to zero' in Figure 44' with the pole as its vertei and the polar axis as its initial side, as along the terfor a positive number r, if we measure r units from the pole the ordered pair minal side of 0, aunique point P is determined, and we call units from (r,0) the polar coordinates of P' lf r is negative' we measure lrl 45 ifi. i"f" Jn the ray diretied opposite to the terminal side of 0' In Figure we iilustrate the points (2, nl6) and (-2, Figure 44 Figure 45 nl6)' ffi;ffi$rxm;er$ rr' Pole (-2, A Polar axis Inthemannerdescribed,theorderedpair(r,0)determinesauniquepointP gghr many inihe plane. On the other hand, if we are given P there are infinitelyhaving angles many infinitely are there for it, since ;;i* "i folar coordinates For example, 12,nl6), (2,13n16), {2, -llnl6), and terminal side through P. ic,r:um ; (-2,7n16) represent the same point' (Verify this') system so If we ,op"ii*por. a rectangular coordinate system on a polar the polar with x axis positive the and pole tt ai tt e origin coin"ides with ihe and rectangular between relationships axis, as in Figure 46, we can determine and system rectangular the p y) in polar coordiiates. Let have coordinates lx, (r, 0) in the polar system. The following relationships hold true: i I \J ; rtu tan #-.'-;$r d fli Y:rsin Figure 46 ints um rM (x* 12:x2+ v' $:fCOS ol srrc,h dxe[ IIE rystem* catled rhe 0:/x eJ '1 (18) (1e) 366 8/ANALYTTcAL rRrcoNoMErRY The next two examples illustrate how we can use these equations to change from one coordinate system to the other. EXAMPLE 29 a. b. Change(..,8,_-1)topolarcoordinates(r, g),wherer>0and Change (- J2, 3nl4) torectangular coordinates. 0<0 <2; Solution a. From equations (18) we find that -1 tanU:..: .u/ 3 -2_A b. Since x > 0 and,r' < 0, we knox' that I is in quadrant The polar coordinates are therefore (2,llnl6). From equations (19) IV, so 0 : lh b : - -, : cos;: -vr( #) / 1\ ,sin7- -vrftr) : -1 - -\ r.'3; ., 1 So the rectangutrar coordinates are (1, EXAMPLE 30 a. b. Solution a. - 1). Find the rectangular equation of the curve whose polar equation il r: 2 cos 0. Find the polar equation of the curve whose rectangular equation )'2 : rs 2s' For r + 0 rve replacqcos 0 _by xf rto L\ \\X PL obtain ,':2; 12 Now we use the fact that 12 : -2x x2 + y2 to get the desired rectangultr equation: x2 + !2:2x which we recognize as being a circle. The fact that the origin is on th* circle assures us that no further points on the graph are obtained if r 1;-, 10,'POLAR r - 0 is the pole, which coincides with the since the only point for which .i to change COORDINATES 367 origin. b. From equations (19) we obtain cos o -':::j ;::2r r(r sin2 , r-0 | r }cos0csc20 (0 + Zkn) The solutio rr r : 0 is contained in the second equation. When 0 - ii l. the equation r - 2 cos 0 csc2 g gives r -- 0, so r - 2 cos 0 cscz 0 contains all points on the curve. Thus, the answer is r :2 cos A csc2 0 To graph a polar equation we can make a table of values by assigning values to 0 and calculating r. We typically assign to 0 the radian measure of special angles we have studied, and we use our knowledge of the behavior of the functions as 0 varies from 0 to 22. As an alternative, the graph is sometimes more easily obtained by changing to the corresponding rectangular equation. However, some of the more interesting curves in polar coordinates have corrcsponds to complicated rectangular equations. The equation r cos 0 in rectangular coordinates, so we know that its graph is a vertical line. In this case it is clearly easier to work with the rectangular equation. On the other hand, the equation r2:4cos20 corresponds to xa +2x2y2 + y4 4x2 + 4y2: 0. (Verify this). If you attempt to draw the graph from this rectangular equation, you will soon be convinced that the polar equation is easier to work with. In the next example we show its graph. :2 x:2 - E,XAMPLE 31 Draw the graph of the equati on 12 - 4 cos 20. 0 < 0 < z, since cos 20 will complete one cycle in this interval. Writing the equation in the form r: t2Vcos 20,we see that values of 0 for which cos 20 < 0 must be exciuded. Thus we cannot have nl4 < 0 < 3nl4.We make a table of values as follows. Notice that we choose values for 0 so that 20 is an angle whose functions we know. The values of r are rounded to two decimal places. Solution It is sufficient to consider lln I 12 : --l ,1,f'F: 11 1 15 -;]u l!l _i. The graph is shown in Figure 47. The curve is called a lemniscate. 368 8/ANALYTICAL TRIGoNoMETRY Figure 47 lltr 7T t2 n 7t 0 I X: oilt{ Called a r., of heal We now present a summary of some standard polar curves, showing their equations, their graphs, and their names. Straight Lines 2 Line through pole making an rcos 0 - a Same angle a with polar axis asx: a rsin 0 - Same b 6y - b Circles r-Zacos0 T:A Circle'of radius center at pole a, Circle of radius a, center on polar axis, passing through pole x0 POLAR Limagons: r-a+ b cos 0 (a- b, 0) a> A:b r - a(L + tklr in, = r, a<b b Limagon without loop (If a > 2b, the "dimple" on the left is absent.) cos 0) Called a cardioid because of heartlike shape ring Lemniscate z Rose Curvesz 12 r COORDT\{TES 369 : - Limagon with loop o2 cos 20 acosn0 -I -,I & =[ n even (n : 2) n odd When n rs even, there are 2n petals;when , {e* trit" p6ilR r (n: 3) is odd, there are ru petals. You may have noticed that each of the equations we have given involves the cosine function only and that the are all symmetriJ to the polar "urn"Jthat if cgs axis, since cos(-0): cos 0. It can be shown 0 is replaced by iin 0, the effect is to rotate the graph in a countercloctwise direction by 90". More generally, the graph of r : f(0 * a) is the sanre as that of r : /(0) rotated through an angle a. Since cos{0 - nl21 - sin g, this confirms that replacing '- cos 0 by sin 0 rotates the graph rf 2 radians, or 90". Similarly, since cos(g -cos 0 and cos(0 - 3n12,1: -sin 0, replacing cos g ") by -cos 0 or by _sin J 0 370 8/ANALYTICAL TRIGoNoMETRY In the case of cos n0 being teth" rotation is 90"fn, with similar results for the other rotates the graph 180' or 270o, respectively. placed by sln i0, por example, the graph of the equation 12 : 2a sin 20 is the lemniscate rz :2acos 20 rotated tluough an angle of 45". We conclude this section with an example of a type of problem often encountered in calculus' cases. EXE RC H,XAMPLE 32 r: 3 cos 0 and r: 1 + cos 0 on the same polar coordinate the area inside the first curve and outside the second. Also shade and system, find all points of intersection of the curves. Draw the curves a Solution The first curve is a circle of radius ! centered at @, 0); the second curve is the in one described is the area cardioid, as shown in Figure 48. The shaded problem. To find the points of intersection we set the two r values equal to one another and solve for 0: 3cos0:1*cos0 2cos0:l 1 cos0- 2 7T 5n e- 1,7 Figure 48 at(1, nl3) and (|,5n 3tis probably better to give the equivalent representation (2, -rrl3) for the We flnd r to be It second point. * in each case. So the curves intersect ny s&mg the two equations simultaneously we found two points of intersectlon. Buifrom the giaph we see that the curves also intersect at the pole' Why did this not show up in our solution? The answer lies in the fact that rrr hqimg rB* b ffistr mflffirrumr8 fiffi] PoL{R c(loRDI\{TEs 371 the curves pass through the pole for different values ol t. The circle g.'o through ttre pote wtren g:n12, whereas the cardioid -errs throrytr rire poie when 0 : zr. Nevertheless, the pole is a point of intersection' lm'ttliff; EXERCISE SET IO AlnProblemsland2plotthepointqinapolarcoordinatesystem. uu&iml,mmr d_ mr8 l$ / 3z\ e' (-2,/ '' e Ed iEE fu qIBe- 1^('19 bt-;) c(-,+) d('il 2. a. liCI e' 3. (4,n) / b. (-r,) c. t,-?) d. (o il 5z\ [-'' - r/ Give flve different sets of polar coordinates for the point (2,2n13). lnclude at least one set for which r < 0 and one for which 0 < 0. In problems 4 and 5 give the corresponding rectangular coordinates for the given polar coordinates. b(-*';) 4.a('':) / dk') c(^') :)z\ e' (.-t,-;J 7z\ h. / s. a. (r",r.-Zl / -9n\ e. (Vt,Z/ fr.lq \ 6/ c. /-,0, \ -T) 4/ In problems 6 an|T glvethepglar coordinates (forwhich r>0 d. and @,3n) 0<0<2n)cote- sponding tq the points with the given rectangular coordinates' c. (0,5) d. (-4,0) 6. a. (1,.F) e. (-3, -J3) q. G2,2) 7. a. (4, -4) e. (4,try, -J6) b. l-2.f,2) c. (0, -2) d. (; +) In Problems 8-18 change to the equivalent polar equation' ,a * t'ff. )m n, --'" h for *ffi of u:lm- fu hEt ptnrue lirn'lt 9. x+Y:O 8. x?+y2:9 10. 2x - 3Y: ! 12. y+l:0 14. x'-Y':4 16. y2 :2x + | 18. ,tnT -2: 11' x:3 13. *'+Y'-4y:o 15. x!:2 l'1}. (x * 1)2 * -i'2 : x 1 372 8/ANALYTICAL TRIGoNoMETRY In Problems 19-30 change to an equivalent rectangular equation. TE 19. r-4 20. 0-- 21. rcos0:2 ?3. 2rcos0-3rsin0-5 ?5. r:3 sec 0 27. r- -2cos0 22. rsin0- -1 24. r :2 sin 0 26. r : 4 csc 0 28. 0 _ tan-t z 29. va_ I 30. l-cos0 3 r2 sin 20 : I In Problems 3l-52 identify the curve and draw its graph. 3L. r :3 33. 0:. r: 536. r:4 cos 0 r:I*sin0 r - 4 - 5 sin 0 r:2cos20 r - -cos 30 12 - cos2 0 - sinz 0 r :2 cos 6 - -) 34. rcos 0:2 3n 4 35. r- -3csc0 37. r: | - cos 0 39. r - 5 - 4 cos 0 41. (' : 9 sin 20 41". r-sin30 45. r:3 sin20 47. r :2 * sin 0 / \ B a) J-at- 38. 40. 42. 44, 6. 48. 3z\ tt 1 49. r-2cosl0--l so. r- 2+2.".(r.;) 51. r - 3(sin 0 - 1l 53. Write in polar forrn. 52. 12:8singcosg a. rr-1., _ 4/ 4{_Tl * l.r}3 r 54. Write in rectangutrar fonn. 4 L' r- L-zsino b' Draw the graphs in Problerns 55. b. *, - y, 55 - 4x coS 0 + 0 r sec' 0 - sin 0 - 3 cos 0 -61. 4 r:2rrn'l 57. / - - 2+cos0 sin 0 Hint. Write in the form r : a cos(O - u) 58. l: 3cos0 -4sin0 (SeehinttoProblem5T.) 0 59. "z r=2cosi OO. ,:Jqsirr0 61. r:3cos59 62. Bychangingtorectangularcoordinates,showthatthegraph of r:4sec210121 isa parabola. Draw the graph. Problpms 63-65 involve threq spiral curves. Dtaw them for the indicated values of a. 63. @. Spirpl of'Archirnedes, Hyperbolic spiral, r : r: a0 (a: al9 (a:2) l) 1I/PARAMETRTC EQUATIONS 65. 66. Logarithmic spiral, r : eoo (a - l) Graph the bifolium r : a sin 20 cos 0 for a - 373 1. In Problems 67-71 show the two curves on the same coordinate system, indicate the specified area by crosshatching, and flnd all points ofintersection. 67. Outside r :2 and inside r : 2(1 + cos 0). 68. Outside r: I * sin 0andinsider:3 sin0. 69. Inside r : 3 cos 0 and inside r :2 - cos 0. 70. Inside r : -4 cos 0 and outside r : 4 *4cos 71. Inside r : 5 cos 0 and outside r : 2 + cos 0. 11 0. Parametric Equations As we have seen, some curves are more conveniently represented by polar equations than by rectangular equations. There is, however, still another way of representing curves by equations that in some cases is more natural and is better suited to analysis of the curves than either rectangular or polar equations. In this third way,a rectangular coordinate system is used, but x and y are each written separately as functions of an auxiliary variable, called a parameter.ly assigning permissible values to this parameter, points (x, y) on the graph can be found. For example, consider the equations [x:t-l :2t LY ,." where I ian be any real number. Here the letter r is the parameter, and the two equations together are called parametric equations. To draw the graph of the curve represented by these equations, we can construct a table as shown below and then plot the point (x, y) on a rectangular coordinate system (Figure 49). The graph appears to be a straight line. Note that the parameter r does not appear on the graph. We use it to f,nd x and y but plot the points (x, y) only. Figure 49 ntg'-r mda- ma 37 4 8/ANALYTICAL TRIGoNoMETRY is desirable to find the rectangular equation of a curve whose parametric equations are given. In our example, we could solve for t from the : first equation, getting t: x I 1, and substitute into the second, getting y In the examples line' is straight graph a 2x + i,which confirms the fact that the that follow, we will see some other techniques for eliminating the parameter. but it should be noted that it is not always possible to do so. It is important to be aware also that the parametric equations may impose certain restrictions on the graph that are not present in the rectangular equation. We will illustrate this in some of the examPles. In general, parametric equations for a plane curve C are of the type Sometimes it (x: f(t) tY s(o : varies over some prescribed interval I (which may be all of R). The totality of points (x, y) obtained by letting , vary over I is the curve C. When C is described in this way, we say it is represented parametrically. The letter r is commonly used as a parameter, since in applications it often represents time. However, other letters may be used. The next three examples further illustrate where , these concepts. EXAMPI-.,E 33 Let C be the curve defined parametrically by (x-2r \;:; -23r<2 Eliminate the parameter and draw the graph of Solution From the first equation we get t: C. xl2, and on substituting into the second, we have x2 :4y which is the equation of a parabola. We note, however, that with the given domain for t, the graph extends only from (- 4,4) to (4, 4\, as shown in Figure 50. Figure 50 EXAMPLE 34 Show that [*-2cost tr-2sin t is a parametric representation of a circle. o<t<2n EXAU r1/PARAMErRrcEeuATroNs 375 rwhosg rmrk [8 -!i Solution To eliminate the parameter *' ,: : 4 cos2 t + 4 sin2 t : 4(cos2 t -f sin2 t) : 4 : rflmpnff or x2 + y2 4,which is the equation of a circle of radius 2, centered at the origin. is instructive to observe how the circle is traced out as , varies over [0,2n]. (*, y) as a moving point whose position is deterWe can think of the point P(r) mined by r. We see from the given equations that P(0) (2,0), P(nl2): (0,2), P(n): (-2,0), P(3n12): (0, -2), and P(2n): (2,0). Thus, the circle is traced out in a counterclockwise direction, starting from (2,0). In this case we can see a geometric interpretation of the parameter r. As shown in Figure 51, it is the radian measure of the angle in standard position with terminal side 0P. rlmttr- It portefi[ rictious hrstru":r fL Tbs + y' we square both sides of each equation and then add: : : Figure 51 - Whsm lsmcr r h titrBg- hsmu.re EXAMPLE 35 Discuss the curve C defined parametrically by J.-1+cos2r [l' d-w,e - sin r 7E 2 TT -2 Solution Observefirstthatforrinthegivenintervalwehave0< x < 2and - 1 < y< 1. To aid in identilying the curve, we eliminate the parameter, making use of the identity cos 2, : | - 2 sinz t. Since / : sin ,, we get E x: CIT{mr rFrymmu I f cos2t:1 + (1 - 2sin2 t):2-2y2 of, y2:-l@-2) This is the equation of a parabola with vertex at (2,0), opening to the left. But because of the restrictions on x and y imposed by the parametric equations, the graph of C consists only of that portion of the parabola shown in Figure 52. Note how the curve is traced out as , varies from -nl2to nl2. Figure 52 376 8/ANALYTICAL TRIGoNoMETRY As we have seen in Examples 33 and 35, the graph of a curYe defined parametrically may be only aportion of the gtaph of the corresponding tectangulat equation. Restrictions on x and y may come about because of the interval for r (as in Example 33) or by the equations themselves (as in Example 35). The important thing is tirat you must take such restrictions into account when using the rectangular equation to draw the graph. The next example illustrates how parametric equations sometimes arise naturally when describing the motion of a moving object. EXAMPLE 36 Solution A projectile is flred at an angle of inclination e (where 0 < a < nl2) at an initiatr spied-of uo feet per second. Find parametric equations for the path taken by the piojectile. Neglect air resistance, and assume the ground is level' Choose axes as shown in Figure 53, and let (x, 1) be the coordinates of the pro' jectile after , seconds. If there were no gravity, the projectile would simply continue along the straight line ofits initial velocity vector and be at a distance ro; on it at time r. But gravity causes the t'coordinate to be diminished by tgt2 after f seconds. So from the figure, we see. using right triangle methods, that [= - ( r'or cos 7 Lj'-r'orslnz'- ist' ) These, then, are parametric equations for the path. We must have t 0, and t is also limited b1 u'hen the projectile strikes the ground. You will be asked t" find this upper limit on r in Exercise Set 11. Also, you will be asked to find the rectan-sular equation olthe path. as well as the maximum hor2ontal and vertica distances for the projectile. l\ Figure 53 l..-r----->l Finally, we derive parametric equations for an important curve known as the iou can think of a cycloid as the path traced out by a point on the circumference of a wheel as the wheel rolls in a straight line along a level path. Ftrr example, a point on a bicycle tire would trace out a cycloid as the bicycle roltri along. The iesulting curve is shown in Figure 54, where a is the radius of tha wheel. Note that the curve is periodic, with period 2na. To get the equations, refer to Figure 55, where we show a typical position ol the wheel, with the point P that traces out the cycloid having originally beer coincident with the origin. Then, since the length of the arc PC is equal to th* cycloid. 1|IPARAMETRTC EQUATTONS hod pare- 377 Figure 54 rtangunm-r fierral fon n 35t The - 4rra frEm urilqg Iffi erise :4a iniufle'l rcm b]-tM - 2ra ) iro 4na Figure 55 f the prclnpily c&m- length of the segment Oe (why?), we have knm r*r W after lrt x -Oe - m - fC - trE - ad - asin @ and y:K,:AC-m-o-ocos@ So we have > 0. emd : {f* : :ar,ked to l.y r find thr o(6 - sin @) a(t _ cos {) -oc<d<oo for the parametric equations. A fact of historical as well as practical importance is that the cycloid is the "curve of quickest descent," in the following sense. Let A and B be two points not on a vertical line. As shown in Figure 56, suppose a wire joins ,4 and I and a bead is placed on the wire at .4. Among all possible shapes into which such a wire can be bent. the one that causes the bead to reach B in the least time is part of an arch of an inverted cycloid, with the origin at ,4. This is rather surprising, since one might think a straight line would be best, since this is the d veutilcai shortest distance. The curve of quickest,descent is called a brachistochrone. The mt as discovery that the solution to the brabhistochrone problem is the cycloid was made independently by a number of seventeenth-century mathematicians, including Johann Bernoulli and Blaise Pascal. In the exercises you will be asted.to find the rectangular equation of the cycloid. You will discover that graphing thp curve directly from the parametric equations is easier than using the reitangular equation. &e r the cir- nth For p* ro,fl3m m of thu uifiom of tBy hffi rk rI to Figure 56 378 8/ANALyrrcAL rRrGoNoMErRy EXERCISE SET Review Exe 11 A In Problems 1-6 draw the curve represented by the parametric equations by assigning values to the pararneter to determine points (x, y). l. (x:2t-l { ["u:1*2 3. (x:2-t 5. - lt:2, : sin2 t [y:cost 2. (x:t2 { [y:r-l 4. fx:sint --oo<f<co 6. fx - tl: (x: cos -co<r<oo t ,lt - '1 Ly:t-Z In Problems 7 -20 draw the curve represented parametrically by first noting any restrictions and then eliminating the parameter to obtain the rectangular equation. 8. (x:3-4t --'l!:3t-l 10. (x:2cosr -.-' 0<t<2n { ' ly::rirrt 12. 1fx:sinr-t ft ft 2="-2 [y:cos2l-1 --<r<14, (x: e' ly:r" 16. fx: ht U: , 18. lx:2tant < 0<t<-- -" -2 [y:.ott 20. _LasaL {x:ztan0-3 2-"-2 [y:3secg+Z 7. lx:2t-l U:1*2 9. [x:cosr-l 0<t<2n i [y:sinr+2ll. [x:cos2t U<t<ft < iY:cort 13. fx: sec t ft n -- -2 2 [y:tanr 15. (x: e' [-t': e-' 17. [.t:r] { , -:t_<t<:r{.t':t' le. o<0<2n {'t:lsin0-l [1:4cos0*2 7L B 21. Find the rectangular 22. equation of the path of the projectile in Example 36, and identify the curve. For the projectile in Example 36 find: (a) the time at which the projectile strikes the ground, (b) the maximum horizontal distance covered by the projectile, and (c) the maximum height the projectile attains. 23. Find the rectangular equation of the cycloid. 24. Draw the gra-ph of one arch of the cycloid with a : I 2.Take { in the intervalf0,2n}, and use a calculator to obtain points. 25. Draw the curve defined parumetrically by n - cos3 0, y sin3 0, for 0 < 0 < 2r. Find the rectangular equation of this curve (the four cusp hypocycloid)" 26. With the aid of a calculator, make a table of values for r in the interval [0, 2nf, and draw the graph of the curve defined by f* - t, : f / sin r sin r - r cos, cos r (This is called an involute of a circle.) RE\-IEr,t E\ERCISE sET 379 Review Exercise Set A rmig[Imm'S 1. Evaluate each of the following without the use of tables or a calcularor: 2. Express each of the following in terms of sin 0 or cos g. I a. tan(7n112) a. .-(? - ,) 3. T '; . -, m* I !t! sin(23n112) b. c. * n) e' *'\7len* \ ', cos c. ,nn(o h il. rft#m h!*s" lglun rlm - o) r::H![]*x b. + + p, T,si?iild '".(il a. sin(a B) cos(a c. tat(a - B) g. lf tan 20 :- _! and nl2 < 20 < z,B) flnd sin g and cos 0. 9. If tan 0: f and n <0 <3nl2,findsin(012), cos(012), artd,tan(012). 10. Evaluate: 11. Evaluatb: a. cos[arcsin(-$)] b. a. tan[2cos-1(-a)] b. sin[2 sin-1(-:r)] cos(]tan-1 4.f) In Problems 12-25 prove that the given equations are identities. 12. csc0-cos 0cotg-sing i,14. hlrfu sec(4n cos{t3z l3r Evaluate: 13. 'cos n'trm (z/B) d. a. arccos(-|) b. tan-r(-r,,6) c. arcsin(uTl21 d. cos-11-l,; e. tan-1(-i1 4. Ifsec 6 :f, and sin o < o, o < o 32n, find a. cos20 b. tan20 c. cosf0 5. Ifsin 0 : f and tan 0 <0, 0 < 0 < Zn,fmd: a. sin20 b. cos20 c. sin $0 r mmflH* m d. .or/e - 1) \ 2/ b. 0+cot0:coSg A 1+csc0 16. "'!.'^ !3*: eos 20 thn0 tan] r sec0-1 2tan0 lU. rv' sec0+1' tune -sec0+l 20. sin0*cos 0cotg:csc0 zz. r+-3n2e secg+1-sec0 t^* e aa' 24. L! 1-.unro:sec2o cote 15. L* -' -:srn2e tan0fcot0 cos0 secO_ tana g _l*sinO tan 0* cos 0 17. sin 19. tan g(co s 2e sec 6 + l) _ sin 20 21.. sin20*(cosg-sin 0lr-l 23. cot7-tan7:2cot,20 )F 25' srn 20 1-cos2e:coto In Problems 26-31 find all primary solutions. 26. sinx-2sin2x:0 28. sin2o*cosg:o 30. 2cos23t_ 3cos3t- 2:0 : 27.3tan3x-tanx-0 29. coszo-sing:o 31. 2cos2t:r-sinr 380 s/ANALYTICAL TRIGoNoMETRY In Problems 32-34 sketch the $raphs' b. Y:3cos2x 32. a. y-2sinx*l 33. a. 34. y:ztanff b. Y:2cosT*' b. a. y -rir(1, .;) 35. If / ,, :2(cos flnd zr' Y -2 cos("'-, I 4n+isinrT . 4z\ 5r\ and zz: 3(cos, i"i"l) i+ 5n zrand zrfzr.Express ans\lers in rectangular form' In Probleins 36 and 37 expand using De \{oivre's theorem' 37' 1z+zi,flln 36. (t - i)8 38. Write in Polar form' a. -t2 + 8,r : 16 b. -tr -.1'r ::Y*r+y, 39. Write in rectangular form. a' .t f-_ b' =- --' l--icostl In Problems J0 and 'll r1 cos20:4 dras' the graph' {). a- r:}--1 cos0 b. 12:-4cos20 b. r:2-Zsin0 41. a. rrcosl0:0 In Problems 42-45 draw the 42. 44. I*:,--, graph, making use of the rectangular equation' ti:,;, In Problems 47 -52 sing 50. tan-1 ? csc2 0 - lt lt -i<o<' -co<r<oo ] + tan-t(-z) prove the identities' cos2g--1 _tan. tLAfra sec 0 csc 0 tr:'une {l:t+,' r !1" 4g. Drrr r*2-tanz0-1sin2 w' sec'0 cosu -sin20 sec0-1 49. -- ^-:sec0+1 {x:1+sec20 45. lx:2-e2' 0<t<2n o<t<2n Jx:l-2sinr lir:3cosr B 46. Evaluate: ;. los[tan-' fi - sin-1(-$)] b. 47. 43. -l<r<oo - 2 cot0 csc e - | zcos 0 csc 0 - tan 0 - cot sin 8o - sin 69 cosg0+cos60-vdnl 0 p^ 52' cos o 1-sin0 =tan0+sec0 eosz1 RE\-IE\\ EXERCISE SET 381 In Problems 53-56 find all primary solutions. 55. tan22x*3sec2x*3:0 tan20 :2 cos 0 57. Draw the graph of: 53. /n\ \r/ a. !:3-Zcosl)-2xl 54. cos20ser0*secu:l 56. 1 + sin f : cos r Hint. Square both sides- b. y:J3cosx-sinx 58. Findalleighthroots of 1. 59. Find all cube roots of -64i. 60. Find the complete solution set to the equation x6 + I : 0. 61. Draw the curves r :3 - 4 cos g and r:2J3tl - cos 0) on the same polar r) 62. *:(atb)cos0-a*.(f)e Draw the epicycloid for which I r[. ttllt,,, i'i t* ([< r :offi J# c+. ordinate system, and crosshatch the area that lies outside the flrst curve and inside the second. Find all points of intersection. An epicycloid is a curye generated by a point on a circle that rolls externally around a fixed circle. If the fixed circle has radius a and the rolling circle has radius b, it can be shown that parametric equations of the epicycloid are and y :(a+b)sin0-bsin ft:!\, a:4 andb: \b l.Take 0 from0to2n. / Cumulative Review Exercise Set 1. 2. III (Chapters 7 and 8) In each of the following a right triangle ABC is given, with right angle at c. Find all missing sides and angles without the use of a calculator or tables. a. A:30", b:12 c. A:45", c:16 e. b:6O, c :12 b. B:60", a:4 d. a:10, c:20 Without using a calculator or tables find all six trigonometric functions of each of the following. Leave answers in exact form. a. 4nl3 b. -3n14 c. 17n16 d. 75' e. f. l7nll2 g. -22.5' h. (2n - 1)n, ri an integer 3. Evaluate without a calculator or tables: a. sin[2 sin-1(-t)] b. cos[cos-1(-1s,) + ran-1(-3)] c. tan(l)0, where sec A : ) and sin 0 < 0 d. /n\ cscIf - 0]. 990. wherecot g :landcos0<0 \4) 4. Prove the identities: _ a. 5. sec0-cos0+txt9 _,_-n , (sin0-cos0)2 l-tan? b. tand*seco cos20 l+tano Let ZL:+(.o, + .i sin +) and Zz:2 (.", 2+, r,r;) Find: a. -:SmU ZrZz b. ztlzz c. z! 6, 7. d. The square roots ofz, Express answers in rectangular form. Points z4 and B are on level ground on a line with the base of a tower. From the top of the tower the angles of depression of A and.B, respectively , are 24.6" and 31.2. . If points ,{ and .B are known to be 20 meters apart, find the height of the tower. Evaluate without using a calculator or tables: a. cos[2 sin-1(-.1)] b. sin[2 arctan(-3)] c. tan|fcos-1(-fi)] d. cos]a, wherecsc o: -iandn<u<3n12 8. Find all primary solutions: a. 2 sin x * f tan x : 0 b. 2 sin2 lx - 2 cos2 x : l 9. A chord on a circle ofradius 6 inches is 10 inches long. Find: a. The central angle formed by the radii to the end points ofthe chord b. The arc length cut oflby the chord c. The area ofthe sector formed by this arc and the radii d. The area of the triangle fbrmed by the chord and the radii 10. [n each of the following, triangle ABC is a right triangle, with right angle at C. Find all unknown sides and angles. a. A:16.3e, b:4.68 b. B:5240', c:171.5 c. a:13.7, b:21.3 d. b:50, c:130 11. a. A curve on a railroad track is in the form of a circular arc which is 1.5 kilometers long. Ifthe central angle corresponding to this arc is 72", find the radius ofthe b. circle. A girl is riding a bicycle with 28 inch diameter tires which are rotating at the rate of 200 revolutions per minute. Approximately what is the speed of the bicycle in miles per hour? sides and angles of the triangles ABC. Also, find their areas. 12. Find all unknown a. A:32.4', B:47.8', c:25.3 b. B:28.3., b:23.5, c:39.2 c. A:98"24', b:124.3, c:89.76 d. a:32.5, b:26.8, c:18.6 383 384 CUMULATIVE REVIEW EXERCISE SET III (CHAPTERS 7 AND 8) 13. Prove the identities: sin 0 + sin. 0- . -) cot0 a. sec0-1'rec0+l-'vv.. 14. Find all primary solutions: a. sinx*sin2x-0 15. Find all solutions to the equation L6. Evaluate: a. b. c. 17. b. sin[cos-1(-f) * Draw the graphs a. - + 64i x b. cos 4x - 0. - 2 tan(xl}) cos 2x tan-1(-Pr )] arctatf cos(u+2fl), where seca: arctan(-2) x3 stn 2x cos (1 + cos 2x) cos21x121 f, csc fl < 0, cotB:2, secf < 0 : y:r +2sin(;-, b. r:+*'(**i) 18. An airplane leaves an airport at 2:O0 p.tu. and flies at a heading of 148" at an average cruising speed of 240 miles per hour. Another plane leaves the same airport at 2:15 p.rra. with a headingof 25'and cruises at an average speed of 400 miles per hour. If they fly at the same altitude, how far apart arc they at 3:30 p.rr't.? L9. Prove the identities: a. 20. 2cos3 0 *sin20:2cos0 l+sin0 A Prove the identities: sin0 1-cos0 L. cot0srn2e-cos20:I 21. 22. 23. Draw the graphs: a. y - ]tan1zx12) b. ! : ] sin-'(Z*) + nl3 ?7. b. cos x sln x =++th3':"..2, rig in the of Mexico is viewed from points A and B Gulf offshore oil drilling on the shore, with the point on the shore nearest the rig lying between A and BIn the triangle formed by the rig and the points ,4 and B, the angle at A is 76"25 and the angle at B is 68'54'. If A and.B are 6 kilometers apart, find the distance from, each of these points to the drilling rig. How far is the rig from the nearest point on shore? Show the curves r : 2 cos 0 and 12: Z.,6 sin 20 on the same polar coordinate system. Crosshatch the area outside the flrst curve and inside the second, and find all points of intersection. Prove that the area of the parallelogram having the vectors V and W as adjacenl sides is lVl lWl sin 0, where 0 is the angle between V and W. In the accompanying flgure, triangle ABC is in the horizontal plane and line CD is vertical. If side c and angles a, B, ard @ are known, derive the following formula for the length ft of side CD: 24. An 26. e 2 Observer ,4 is stationed at a point due west of a monument, and measures the angle of elevation of the top of the monument to be 45'. Observer B is 100 feet due south of observer A, ard.B measures the angle of elevation of the top of the monument to be 30". Without using a calculator or tables, find the height of the monument. Find the complete solution set of each of the following: a. .r,6sinx:cosx-l 25. /u 7t\ sina-1 b. tanl l-\2 4) cosn o_csinutan0 ttsin(a+p) CUMULATIVE REVIEW EXERCISE SET trII ICHAPTERS - If c:21.34, a: I $en,r -r,- I alrpnot"i .a: I mile= pr: L: 28. I each of the following: 0 I. , ) sin2 , [, -2sin20 b' [. :et*e-t IY :et-e ,rg-ffi fm[IU]ffii: [TEflt tJ&ffJ A eru -* iis*6 1 frIffi:::,,M tre$[ ;ullrr. mCI{rc":,"u"]ii t ad :ns n d,_ra.m:rr [ire Cf s g fbrcrru,iiu - 56.42'. find ri" A,s,.- nrj ,:i-: The pilot of a small alrllafre flew from to'wn to town -B and returned the follorrins {ay. _The bearing of B from i4 is N 42.3' E, and its distance from A ir-+sz^.,rr*. on the trip flrom A to B, a 33.4 mile per hour wind was blowing from 270.: on the return.trip the wind wab ft'om 305' aizg.l miles per hour. If ihe pilot cruised at 182 miles per hour airspeed, find hbr heading on each leg of the i.if. ato find her total time in the air. a' du€ yc,r-ii. and O 38s D 29. Graph tk fr:43"16., ] AI A.n "l-*'ff- $ 18'37', A\D ! t 0<0<n oO<t<Q AN-32 ANswERs ro SELECTED PRoBLEMS Review Exercise Seto page 305 5. A:45"'a:sJl,o:suD 3. A:30',c:8,b:4J5 l. B:30',c:t0,a:514 7. A:30',r: 60",b:7J1 9. B:57",a:7.08,b:10.90 l1.. a: lll.34, A:4',7.51, B:42.49" 13. A:63.8"a:30.06, D:14.79 15. a. 4nl3 h. 1nl4 c. nll2 d. 3n e. 8nl9 17. a. 4n x 12.5'1 h. (2401n)' x 76.39" h. -1 i. -2 i.-J312 re. t. ! b. -+ c. 1l$ d.0 e. -J, r.2 s.t 21. b : 10.72, c :20.82' C: 118.9' 23. Solution1: A:64.49",8:83.51", a:13.62;solution2: A:51.51',B:96'49"a:ll'82 25. A:100.2",8:52.2',C:27.6' 27. Nosolution 29. A:37'36",8:43'05", C:99'59" 31. Solution l: B:104.2t",C:50.59", b:34.38;Solution2: B:25'39",C:129'41",b:15'21 33. 61.2 feet 35. 103.12 feet, area : 4,914.91square feet 37. lvrl : 15.0, lvr + Vrl : 39. lRl : 80.6 pounds, 0: 53"38' 41. 22.86 feet 43' 526'46 miles 45. 8.48 miles from first tower, 7.93 miles from second 47. 646 feet, S 10.6" E 25.9 Chapter 8 Exercise Set 1, page 314 sing- -Zrlltycos 0 - _1,csc0- -312'f1,sec0 - -3,tan 0:2t{2,cot0-lpp sin0- -#,cos 0:*,csc0- -#,sec0-+,tafi0_ -+,cot 0: -* \ sin0- -rlBtO.cos 0:i,cscg- _ 4ltlls,sec 0:4,tan0- -J15,cotg- -U^,lB : tll,cot 0 lltl3 7. 0 - 4nl3; sin 0 - -15 2. cos 0 - -!, csc 0 - -z|",ft,sec 0 - -2,t3n0 g g ztlr,cot 0 tan 3, 0 csc 0 cos sin 0 9. -j. == -Z.rlitl" - - UztD= - - -3l2tlr,sec (€. (0,-1) d. (c. 1. 0t (0, b. 1) +, - ,ll lz) 11. a. eJ3p,+) 1. 3. 13. I IX r9 ?1 d. a. v 3,1 3? a. nl6 b. nl4 c. n 3 d. nl2 e. 0 . a. sin 0.3 : 0.2955, cos 0.3 : 0.9553, tan 0.3 : 0.3093 b. sin 2.718 :0.4110, cos 2.718 : -0.9116, tan2-'ll8: -0.4509 c. sin 14.03 :0.9943, cos 14.03 : 0'1070' tan 14.03 :9.2955 cos( -3.625) : -0.8854, tan(-3'625): -05249 d. sin(-3.625):0.a64S, -0 c. nl4,5nt4 d' 0, z e. nl2,3nl2 f' 0,n g' 3nl2 h' z i' 3nl4'7n14 19. a. nl2 b. 21. a. p(g)iseither(0, 1)or(0, -1),sotangisundefined b. P(g)iseither(1,0)or(-1,0),socotOisundefined c. piei iseither(0, f) o.iO, -ti,sosecgisundefined d. P(0)iseither(1,0)or(-1,0),socsc0isundefined 23. IlkeR,tan0:ylx:/c,with(x,y)ontheunitcircleif x :tl"tTlz'arldv:t luTi'''Sdif P(0):tttJTTP' 15. 11 klJl +i.2), then tan0: k. Exercise Set 2, page 321 1. cos 0 : -f , csc 0: -2, sec 0: -*, tan0 :1,cot|:1 3. sing:f,cos 0: -l,csc0:i,secg: -3, cot0: -Z 5. sin 0 : _llJi,csc 0: -16, sec0 -- JS1Z,tan0: -i,cot0: -2 7. cos0 = -Z0fi,cscg:3, sec0: -3l2J1,tan0: -llzJr,cot0: -2Jl Exercise Set 3, page 327 3. a. -d6 + Jztrt- b' -(..6 + ,fzttt nllltq 't. a. (r,6 + "Dltq b. (",6 + Jluq e. o. -# "l (-\tr + 4,f2)P 13' a' ttlrtJs b' 3slr7J5 ls' i. -3t tr. a. QJL| -z)le b. cos(nl2) sin 0: cos 0 cos 0 sin(z/2) + e): sin(nlz a. + 17, b. cos(z/2 + 0): cos(nl2) cos 0 - sin(z/2) sin 0 : -sin 0 r. s. \Jz + J6l4,tJ1- t.,|z Jatn -.'Gy+ t. -t.-6 + b., -* b. -ffi ANSWERS 19. a, b. sin(z cos(z -* 0):: 0) - sin n cos 0 coszcos 0 31. 0,n 33. a. sin acosB b. cos oc cos z sin 0: To SELECTED PROBLEMS Af{-33 sin 0 + sinz sin0 : -cos0 cos 1l + cos d sinB cos ? cos Bcos 7 sin a sinf cos y - + - cos d cosB sin y sin a cosf sinl - sin a sin B sin y - cos d sin f siny Exercise Set 4, page 332 1. sin20: -#,cos20: -* 3. sin20:ffi,cos2e:# 5. sin20: -f,cos20:t 7. sin20:2xJT-7,cos20:t-2x2 9. sin0:"tjtZ,cosg: -.,fq: il. sin e: -i,cosg:t B. t JI -1112 b. {z a ,1t12 c. uD + Jttz d. {I -J12 15. sin(a/2):3lJ1.:,cos(al2): -21\ry, fi. sin(al2): -Jttt,cos(al2):,f613 19. p(0):(i,+) 21. a. -sin 0 b. -sin 0 c. cos 0 d. -cos 0 e. cos 0 39. sin 20 : 2xlQ + x2), cos 20 : (t _ x21l1t + x21 41. sin30:3 sin 0 -4sin3 0, cos 30:4cos3 0- 3 cos 0 Exercise Set 5, page 336 1. a. 2+rll b. 2-J3 3. + 5. . tan 2a - #, tan 2fr : -#, tan(af 27 - -1, I l. a. t'612 b. -,l6tz 13. a. O2 7 (e sQlp - 4 9. tan20 - -+, tan(012) : 3 b. - rl2tZ 15. a". 2 sin 4x cos x b. + tan(Bl2) j(sin 8x * sin 2x) Exercise Set 6, page 343 1. {nl6,1tnl6} 3. {nl3,5nl3} s. {nl3,nl2,3nl2,snl3} 7. {nl3,n,snl3} s. 11. 19. 23. 27. 31. 37. {nl2,3nl2} {nl6,nlz,5nl6,3nl2} 13. {nl3,2nl3,4nl3,5nl3\ 15. nos-olution 17. {nlz,3nl2} {0,2nl9,nl3,4n19,2nl3,8nl9,n,r0nl9,4nl3,1,4nl9,5nl3,t6nl9} 21. {O,n12} {n124, 1tnl24, 13n124,23n124,25n124,35n124,37n124,47n124} 25. nosolution {2nl9,nl4,5nl9,3nl4,8nl9,rtnl9,5nl4, l4nl9,7nl4, l7nl9} 29. no solution {0.848,2.29,3.87,5.s5} 33. {0.464,r.Lt,3.6t,4.2s} 3s. {0.285,2.36} {nl3,2nl3,3nl4,4nl3,snl3,7nl4} 39. {nl6,snl6,3nl2} Exercise Set 7, page 352 1. Amplitude - 1, period - z 3. Amplitude : 2, perio d - 2n 5. Amplitude - 3, period - 4 7. Amplitude : 2,pertod - 2nl3 9. Amplitude * l, period :4n 11. Amplitude : 1, 13. Amplitude - 2, penod - 2nl3 15. Period - period - n 1 \ t.it\ AN.34 ANSwERS To SELECTED PROBLEMS 17. Amplitude - 1, period - 2n, phase shift 21. Amplitude - 2,penod - 2n13, phase - nl3 shift -- -? 19. Amplitude - 1, period : n, phase shift : -nl4 23. Amplitude - 2, penod : 2, phase shift :i \ 9. v 2 11. 2\. 23. 25. 27. 25. Amplitude - 2,pertod - ?m 3. phase shift - 29. nl6 31. 33. Jf. 37. 27, a. y :.@ sin(x - z/41: amplitud, : Jl,period : 2n, phase slttft : nl4 b, y:5 sin(-x * 0). where 6 : 2.5 radians as shown;amplitude: 5, period 39" 41" :2n,phase shift: -0 x -2.5 45. +' ,Bn - Exert 29. Period Exercise Set : 6n 8, 1. page 357 1. a. -nl3 b. 5nl6 3. a. 3nl4 b. nl2 5. r. 0 b. -nl6 1r. a. t b. i 13. a. zJitz b. -+ s. a. -2A13 b. i le. -#g 21. + 23.+ 2s. 7. a. nl2 b. ls. 2l"E ryrm n. sing:-!,cos0:f 29.* 3r. 17. 7nl6 -X 1+ 5. 7" 11. Exercise Set 1. 0 9, page 363 21" -7nl4,r: O;l - - 'fiTcos(7nla) * isin(77114)1 i 3. 0: llnl6,r:2;.,/3 - i - 2[cos(llnl6) * i sin(ltnl6)) 31. ANSWERS TO SELECTED PROBLEMS 5. 0:fi,r:4; -4:4(cos n* 9, 0:0,r:5; 5:5(cos 0 + z 7. 0 - i sinzr) 4n13, t' :4; -2 - 2irfr - 4lcos(4n13) AN.35 + i sin(nl3)f sin 0) 11. -t - i"E 13. -si 15. zJz - zi 17. l - i tg. etJtlzl + eilz) 21. zF2: 6Lcos(3n;2) + i sin(31112)): -6i; zrlzr: !lcos(-z) + i sin(-z)l : -l 23. zp2:8[cos(52i4) +, sin(5nl4)] : -+1D - aia;z1lzr:)lsor1-nl4) + isin(_nl4)f : O - iO 25. zlzr:$[ser113n6) -isin( t3r16)]:4\4t4i:zrlzr:t[cos(32/2) +isin(3ltt2)f : _ it2 27. zF2 -24(cos 170' + i sin 170'); zrlzr: l(cos 30' + i sin 30') : (\E/3) + (r/3) 29. 129lcos(9n12) + i sin(9n12)f :729lcos(nl2) + i stn(nlz)f :729i 31. 256lcos(4ht3)+isin(44itl3)):256lcos(2n13) +,sin(22l3)l: -128 + tZtiJS 33. 256lcos(20n13) + lsin(202l3)l :2s6lcos(2n13) + iiin(2n13)): -t28 + nei15 35. (o : 2lcos(nl6) + I sin(n/6)l : Jr + it (r : 2Lcos(5nl6) + i sin(szl6)l : - !5 * i; (, : 2lcos(3n12) + i sin(3n12)) : - 2i 37. (o:1(cos0+ isin0):1;(r:1[9o_s(z/3) + i sin(z/3y] :_i_+iJ1.lz;(z;rlcos(2n13)+isin(2rll3)): -\+i,fZ1Z; (a: l(cos z + lsin n): -1; (+: 1[cos(4213) + isin(4ft13)] : -] - iJllZ; (s: 1[cos(52l3; +;sii1Sz7:i] I - iJ31z 39. (o:4lcos(3n14)+ i sin(32l4)l : -zuD+zi.,,D,h:4lcos(7n14)+isin(7n|4)f :zuD-ziJl 41. {\6 +;, 2i, -,,4 + i, -.lE - i,-2i,,,8 - i} 43. 1z,t + irz, -t + iJl, -2. -l - iy6, r -;nri1 45. By the hint we have p'(cos n$ + i sin nd) : r[cos(0 + 2kn) + l sin(0 * 2kn)]. So p" : r, p : yltn, and n0 : 0 + 2kn. Therefore,O:(0+2kn)ln.Bytakingk:0,1,2,...,n-l,wegetallthedistinctanswers. Exercise Set 10, page 371 2: ku i l. 3. ',, (2, 8nl3), (-2, 5nl3), (2, - 4nl3), (-2, - 7rl3), (2, t4nl3) It rit2 5. t. F] l9r L. (2, 2) b. Q ury, - 3) c. (- s ,12, s Ol d. ( - 4, o) e. (2,2) a. go, Tnla) b. (4, 5nl6) c. (2,3n12) d. (I,5nl3) e. (z.vT,7n16) 9. 0 : rcos0:3 13. r:4sin0 2t. x:2 23. 2x-3y-5 11. Tr 31. 33. 15. 25. 12 -r:3 4lsin20 17. r - -2 27. x2+y2+2x:0 35. 0 19. xz + !2 : 29. yr:l+2x 37 . cas 3nl4 16 cardioid Afd-36 39. ANSwERS To SELECTED PRoBLEMS 43. 41. lemniscate limaEon 4-leaf rose 3-leaf rose (,,?) (r,?) 47. 1: 49. limaEon 51. circle : o,T) cardioid o,+) s (',?) 53. L. r - 55" 16 csc2 20, together with r : 0 b. r :4 r:l-cos0 cos 0 sec20 57. r-Jrcos(0 - 7T14) circle cardioid 11. { : (r,+) 59. 0l0 nl6 rrl4 nl3 r I 2 1.93 1.85 1.73 7rl2 t.4r 2nl3 3nl4 5nl6 Tc 7nl6 5n 14 4nl3 L 0.77 0"52 0 -0.52 -0.77 -1 3nl2 5n l3 7 nl4 -1.41, -1.73 -1.85 Use symmetry llnl6 - 1.93 15. 65. 63. 61. 5-leaf rose I9. .'i"', {X t" 4;t \J.T/ (,,9) 67. 69. (2, nl2), (2, 3nl2) t|, 7T 3t, (* . 5n 3) 71. (2, nl3), (|,5n13) 23, Exercise Set 11, page 378 -1 ',- = Review 0 -s -3 -1 | 23 1. a" 3 4 3. 7, 11. a" a. a. I ANSWERS TO SELECTED PROBLEMS 3. ' t 1 2 x I l. 0 AN-37 . lllf ,* d,i .r.> v u, 7. x - 2y * 7: 9. (x+l)'+(y*2)r:1 0 i {r _41 w4 rl lx Il. x -2y, - I,lrl < 1, ll,l < 1 13. x2-y2:1,x21 - :r!m iffi, .L.lp,:r 15. x!:1,x>0,.1 >0 tg. (, l1)' g ,(y-2)'_ 1,6 17, !2 : 21. x3 ! : x tan a -d;ly!, parabola LUO 25, x2l3 + !213 * | Review Exercise Set, pag e 379 b. i7 - 46 c. J'TA d. -VzEE -V 1-./3 4 4 3. a, 2nl3 b" - 7Tl3 c. *7113 d. ft €. 5. a. _# b. _* 7. a. -H b. Sf c. t* 9. sin(?12) : 5lJ 34,-TEl4 cos(l12) _ _ 3lr[U, ian(L12) _ _ t Ll..a. -+ b. 2lrfi 27. {0,n, nl6,5nl6,7nf 6, llnl6} 29. nl6,5nl6} r. a. tii {inl2, c. 31. 2l"E {7n16, |lnl6, ft12} AN.38 33. ANSwERS To SELECTED PRoBLEMS a. b. Amplitude - 2, perrod : 1l 4 17 35. zt' Zz: 6[cos(l3nl6)* i sin(L3nl6)f : 6[cos (nl6) * i sin(z l6)): 3 J1 + 3ii zrlzz - f [cos (- nlz)* i sin( - 7El2)1 - -2i13 37. 256[cos$nl3)* i sin(4n13)7 : -128 - l28iJ3 39. il. 5x2 - 4y', + 24x * 16:0 b. x2 - !2 :4 41. L, (,*) 43. 23. 45. (y - l)' ''2:x-2 53. {n13,2n13, 4n13, 5n13, 57, a. 59. {4i, -2.18 Amplitude - - 21 nl2,3n12} 55. 2, perrod 2i,2J3 - : 2i} TE, {n12, 3nf 2, n phase shift n /6 61. (3 + zJ3, TT 2), 2* x < 2, y > l6]1 \ /n x): \ 2({cos.x-- jsin x):2sin(;-2 sin (, - ,), \2 plitude - 2, period * 2n, phase shift : nl3 b. 5nl6), (3 : -(x - + zJa,7nl6), 29. pole ChI Exer I, 15" 2!. Cumulative Review Exercise Set III (Chapters 7 and 8), page 383 1. a. B-60,c:816,o-4rl3 b. A-30",c-g,b:4",13 c. B_ 45",a:b:8J, d. A-30",8:60o, b:LO.r6 e. A- B-45",a-60 3. a. -# b. ;+ ". -tlrl5 d. -..,6 5. a. zp2: 8[cos(32l2) + isin(32l2)] : -81 b. zrfz2:2lcos(7n16\ + isin(7r.16)): -.J5 c. z!: l6lcos(2n13) + , sin(21tl3)l : -a + aif d. (o : 2lcos(2n13) + , sin(27rl3)l : - I + i,fr; (, :2lcos(5n13) + , sin(52l3)l : t - inll 7. a. Z b. -+ c. f d. tt,t5 9. a. 1.97 radians b. 11.82 inches c. 35.46 square inches d, 16.58 square inches i 23. ANSWERS TO SELECTED PROBLEMS 11. 17. a, r : 1.194 kilometers b. 16.66 miles per hour a. Amplitude - 2, period : 47r, phase shift - )LILI - 11' J 15. {4i, -2111 b. - 2i, Amplitude 2$ - - j, 2i\ period : 2, phase shift v 21. a. Period - 2 b. a. 25. (rr, ;), (,, ;) 29. b. {0, anl3} a. - -L The given equation is equivalent to * : * sin 2(y or nf 12 < y < 23. AN.39 7rl3), with -7112 < 2(y 7nll2. - 7rl3) < nl2, ,,,r1 {n 6. 5n,6,7n16, llnl6} 27. h _ 1I.76,0 _ 35021' -Xy-2),0<x <2,0<y<2 b. x2-y':4rx>'2 Chapter 9 Exercise Set 1, page 396 t. (4,2) 3. (-1, -3) s. (4,3) 7. (#,?) e. (4,_2) ts. (2,-1,4) 17.(-#,-*) *.(##,t#) 21. If both sides of the second equation are multiplied 8x - 6y - 3 -8x+6y-4 0 -7 (_5,6) 13. (_2,3,4) by - t and then 5 is added to both sides, the result the first equation. So the equations are equivalent. The solution ret 23. 11. i, This is impossible, so the equations are inconsistent. lf -,)' I / 3c - 5\ ,. n}. ) the same as