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CHAPTER 6 SOLUTIONS TO REINFORCEMENT EXERCISES IN TRIGONOMETRY 6.3.1 Radian measure and the circle 6.3.1A. Express as radians i) 36o ii) 101o v) 340o vi) – 45o ix) 27o x) 273o iii) vii) 120o iv) 250o – 110o viii) 15o Solution All we have to remember in converting between degrees and radians is that 2 radians = 360 degrees Then: i) 360o 2 36 = 10 = 10 radians = 5 radians o Note that it is usual to leave radian measure in such form, rather than convert it to decimal form. Once you get the hang of this sort of problem it is little more than practice in cancelling down fractions. 101 ii) Since 2 radians = 360 degrees, 1o = 180 radians and so 101o = 180 radians 120 2 iii) 120o = 180 radians = 3 radians 250 25 iv) 250o = 180 = 18 radians 340 17 v) 340o = 180 = 9 radians 45 vi) – 45o = – 180 = – 4 radians 110 11 vii) – 110o = – 180 = – 18 radians 15 viii) 15o = 180 = 12 radians -1- 27 3 ix) 27o = 180 = 20 radians x) 273 31 273o= 180 = 60 radians 6.3.1B. Express the following radian measures in degrees in the range 0 < 360o. 2 i) ii) 14 iii) – iv) 3 2 3 5 2 5 v) 6 vi) vii) viii) 2 9 4 2 ix) – 5 x) 12 Solution Whereas to convert degrees to radians we have to multiply by 180 to go in the opposite 180 direction and convert radians to degrees we have to multiply by . i) 2 2 180 radians = = 120o 3 3 180 = 7 360o = 360o on reduction to the 360 range ii) 14 radians = 14 180 iii) – 2 radians = – 2 = – 90o = 270o iv) 180 radians = = 60o 3 3 v) 180 radians = = 30o 6 6 vi) 5 2 radians = vii) 2 2 180 radians = = 40o 9 9 5 180 o o 2 = 450 = 90 on reduction to 360 range -2- viii) 5 5 180 radians = = 225o 4 4 2 2 180 ix) – 5 radians = – 5 = – 72o = 308o x) 12 radians = 180 = 15o 12 6.3.1C. Determine the length of arc and the area of the sector subtended by the following angles in a circle of radius 4cm. i) v) 15o 90o ii) vi) 30o 120o iii) vii) 45o 160o iv) viii) 60o 180o Solution The length of arc of a circle of radius r subtended by an angle RADIANS at the centre is 1 given by s = r, and the corresponding area of sector is A = 2 r2 since the radius is the same, 4cm in each case, we can write these as s = 4 and A = 8 and the calculations are then just an exercise in changing degrees to radians. 2 i) For 15o = 12 radians we have s = 4 12 = 3 cm and A = 8 12 = 3 cm2 2 4 ii) For 30o = 6 radians we have s = 4 6 = 3 cm and A = 8 6 = 3 cm2 iii) For 45o = 4 radians we have s = 4 4 = cm and A = 8 4 = 2 cm2 4 8 iv) For 60o = 3 radians we have s = 4 3 = 3 cm and A = 8 3 = 3 cm2 v) For 90o = 2 radians we have s = 4 2 = 2 cm and A = 8 2 = 4 cm2 -3- 2 vi) For 120o = 3 radians we have 2 2 16 8 s = 4 3 = 3 cm and A = 8 3 = 3 cm2 8 8 8 64 32 vii) For 160o = 9 radians we have s = 4 9 = 9 cm and A = 8 9 = 9 cm2 viii) For 180o = radians we have s = 4cm and A = 8 cm2 6.3.2 Definitions of the trig ratios 6.3.2A. Write down the exact values of sine, tan, sec, and complementary ratios for all 'special' angles 0, /2, /3, /4, /6. Solution This may seem like a lot to remember, but with a few aids it is not too bad. It is also very useful to be able to recall these results easily (particularly those for sine and cosine), as they occur frequently (pun intended!) in such topics as AC circuit theory and signal analysis. All you really need to remember are the 30-60 and 45 triangles given in Figure 6.4, and then all of the results are just a matter of definition. Try to get used to using the surds, such as 2 and 3 (See Section 1.2.7). It is much better to leave them as they are until the end of calculation, rather than immediately replacing them by decimal approximations. All you have to remember is, for example ( 2 )2 = 2, which is something like how we handle the imaginary j in complex numbers (Chapter 12), by using j2 = 1 whenever possible. If you didn’t get all the results right first time, use Figure 6.4 to check the following answers. -4- 0 /2 /3 /4 /6 sin 0 1 3 2 1 2 1 2 tan 0 nd 3 1 sec 1 nd 2 2 1 3 2 3 cos 1 0 1 2 1 2 3 2 cot nd 0 1 3 cosec nd 1 1 3 2 3 2 2 (nd means not defined, because in principle we are trying to divide by zero – in practice, however, it is common to regard, for example, tan( /2) as ‘infinity’). 6.3.2B. Classify as odd or even functions: cos, sin, tan, sec, cosec, cot. Solution Again, you either know or don't know that cosine is even and sine is odd. So tan = sin/cos is odd. sec = 1/cos is even. cosec = 1/sin is odd. cot = cos/sin is odd. Remember such results as Odd Even = Odd, O/E = O, O O = E, etc. 6.3.2C. Express as trig functions of x (n is an integer) :i) sin(x + n) ii) cos(x + n) n n iv) sinx + 2 v) cosx + 2 iii) vi) tan(x + n) n tanx + 2 where n is an integer. Solution There is a bit more meat to this question! The safest approach is to use the corresponding compound angle formulae. Also note that we will be frequently using such results as cos n = ( 1)n, sin n = 0, etc, so make sure that you understand these results. i) From the compound angle formula for sin(A + B) we have -5- sin(x + n) = sin x cos n + cos x sinn = sin x cos n = (– 1)n sin x ii) From the compound angle formula for cos(A + B) we have cos(x + n) = cos x cos n + sin x sinn = cos x cos n = (– 1)n cos x iii) tan(x + n) = sin(x + n) (– 1)n sin x = = tan x cos(x + n) (– 1)n cos x n n n iv) sinx + 2 = sin x cos 2 + sin 2 cos x The form of this depends on whether n is odd or even. the easiest way to find the general result is to try a few specific values of n and then see what the general pattern is. We then find that: n n If n is odd then cos 2 = 0 and sin 2 = (– 1)(n–1)/2 n So sinx + 2 = (– 1)(n–1)/2 cos x if n is odd. n n On the other hand if n is even then sin 2 = 0 and cos 2 = (– 1)n/2 n So sinx + 2 = (– 1)n/2 sin x if n is even. n n n v) cosx + 2 = cos x cos 2 – sin 2 sin x = (– 1)n/2 cos x if n is even. – (– 1)(n–1)/2 sin x if n is odd -6- = (– 1)(n +1)/2 sin x if n is odd. n sinx + 2 n vi) tanx + 2 = n cosx + 2 Again we get different answers depending on whether n is odd or even. Using the results of iv) and v) we get If n is even: n (– 1)n/2 sin x tan x + 2 = = tan x (– 1)n/2 cos x and if n is odd n (– 1)(n–1)/2 cos x tanx + 2 = = – cot x (– 1)(n +1)/2 sin x 6.3.3 Sine and cosine rules and the solution of triangles With the standard notation solve the following triangles. i) A = 70o , C = 60o , ii) B = 40o , b = 8, iii) A = 40o , a = 5, iv) a = 5 b = 6, v) A = 40o , b = 5, o vi) A = 120 , b = 3, b c c c c c = = = = = = 6 10 2 7 6 5 Solution i) In this case we are given two angles, A = 70o and C = 60o, and one side, b = 6. From A = 70o and C = 60o we know that the third angle must be B = 180o – 70o – 60o = 50o. -7- A 70 b=6 c 60 B C a We can now use the sine rule to find the two other sides: a b c = = sin A sin B sin C so a 6 c o = o = sin 70 sin 50 sin 60o Hence 6 sin 70o a= = 7.3601 to 4dp sin 50o c= 6 sin 60o = 6.7831 to 4dp sin 50o ii) This is the case of two sides b = 8 and c = 10, and a non-included angle B = 40o. In this case we can get two possible triangles to fit the information given - it is the 'ambiguous case'. Firstly, apply the sine rule to attempt to find 'sin C'. We have b 8 c 10 sin B = sin 40o = sin C = sin C so 10 sin 40o sin C = = 0.8035 to 4dp 8 Now although this looks like one answer, we have to remember that sin (180o – C) = sin C So if we take C = C1 to be the acute angle solution, then there is another solution with an obtuse angle C2 = 180o – C1. Both solutions will produce triangles satisfying the given condition. For the acute angle result we find -8- C1 = 53.46o to 2 dp The obtuse angle result is then C2 = 180o – 53.46o = 126.54o to 2 dp So we have two possible triangles for which to find A and a. A b=8 c = 10 40 B C1 C2 C1 = 53.46o Then A = 180o – 53.46o – 40o = 180o – 53.46o = 86.54o and a 8 o = sin 86.54 sin 40o or 8 sin 86.54o a= = 12.42 to 2dp sin 40o C2 = 126.54o Then A = 180o – 126.54o – 40o = 13.46o Note: If it had happened that the larger value of C produced a negative answer here then the answer would not be a valid solution - we would effectively only have one triangle. So -9- a 8 o = sin 13.46 sin 40o or a= 8 sin 13.46o = 2.9 to 2dp sin 40o iii) For A = 40o, a = 5, c = 2 we again have two sides and a non-included angle. The sine rule gives A 40 b c=2 B a=5 C 2 5 = sin C sin 40o so 2 sin 40o sin C = = 0.2571 to 4dp 5 This gives two values of C, as before: C1 = 14.9o to 2 dp C2 = 180o – 14.9o = 165.1o to 2 dp In the acute case we obtain for B: B = 180o – A – C1 = 180o – 40o – 14.9o = 125.1o But in the obtuse case we see that A and C2 add to give 165.1o + 40o = 205.1o. Since this is greater than 180o no triangle exists for this value of C which must therefore be rejected. For the allowed value of C = 14.9o we find the remaining side from - 10 - b 5 o = sin 125.1 sin 40o or 5 sin 125.1o b= = 6.36 to 2dp sin 40o iv) a = 5, b = 6, c = 7 is the case of all three sides and finding the angles. We can use the cosine rule to find any of the angles. A c=7 B b=6 a=5 C From a2 = b2 + c2 – 2bc cos A we have b2 + c2 – a2 62 + 72 – 52 cos A = = 2(6)(7) = 0.7143 2bc and A = 44.42o Similarly for B we find 52 + 72 – 62 cos B = 2(5)(7) = 0.543 and so B = 57.11o We can then find C from 180o – A – B = 180o – 44.42o – 57.11o = 78.47o v) A = 40o, b = 5, c = 6 is the case of two sides and the included angle - 11 - A 40 6 5 C a B We have a2 = b2 + c2 – 2bc cos A = 52 + 62 – 2(5)(6) cos 40o = 25 + 36 – 60 cos 40o = 15.04373 and so a = 3.88 Now the sine rule gives 5 a 6 = = o sin B sin 40 sin C so 6 sin 40o sin C = 3.88 from which C = 84.02o Similarly we get for B: 5 sin 40o B = sin 3.88 = 55.98o It pays to check our answer by confirming that the angles add up to 180 o (or alternatively of course we could forego the check and determine, say, B by subtraction from 180o). So, the final answer is –1 a = 3.88, B = 55.98o, C = 84.02o A = 120o, b = 3, c = 5 is as for v) but now we have an obtuse angle – but the 1 calculations are basically the same, just remembering that cos 120o = – 2 . vi) - 12 - A 120 5 3 C B a We have a2 = 32 + 52 – 2(3)(5) cos 120o = 9 + 25 + 30 (0.5) = 49 and so a=7 Now the sine rule gives 3 a 7 = = o sin B sin 120 sin 120o So sin B = 3 sin 120o 7 from which B = 21.77o Similarly we get for C: 5 sin 120o = 38.21o C = sin– 1 7 So the answer is a = 7, B = 21.77o, C = 38.21o 6.3.4 Graphs of trigonometric functions Sketch the graphs of i) 2 sin (2t + 3 ) ii) 3 cos (3t – 2 ) Solution Both sin and cosine are of course just continuous waves, and in sketching them we only have to size and locate them. In the general form x = A sin(or cos) (t + ) A gives the amplitude, or the ‘height’ of the wave above the mean level, determines the number of oscillations in a given period of t and fixes the location of the graph by ensuring that at t - 13 - = 0, x = A sin , ie it passes through A sin on the vertical axis, with t plotted horizontally. Using these ideas you should obtain the graphs in the figures. - 14 - 6.3.5 Inverse trigonometric functions 6.3.5A. Find the principal value and general solutions for sin–1x and cos–1x for the following values. 1 1 i) 0 ii) 2 iii) – 2 iv) 3 2 1 2 v) vi) – 1 2 Solution The inverse sine and cosine can be evaluated from the general results (given in radian form) sin–1x = n + (–1)n PV cos–1 x = 2n PV where PV denotes the corresponding principle values, which take the ranges – 2 sin 1 x 2 0 cos 1 x Applying these results, with the appropriate principal values we obtain the table below x i) ii) iii) iv) v) 0 1 2 1 –2 3 2 1 2 sin–1x sin–1x cos–1x PV GS 0 n PV 2 3 2 3 6 GS 2n 2 2n 3 2 2n 3 2n 6 4 2n 4 6 –6 3 4 n + (–1) n +(–1) n n+1 n + (–1) n n + (–1) n - 15 - cos–1x vi) – –4 1 2 n +(–1) n+1 3 4 3 2n 4 where n is an integer 6.3.5B. Find the principal value and general solutions for tan–1x for the following values of x 1 i) 0 ii) 1 iii) 3 iv) –1 v) – 3 Solution Using tan–1 x = n PV where the PV is in the range 2 tan 1 x 2 we obtain the following results. tan–1x 0 1 PV 0 4 3 –4 GS n n + 4 n + 3 n – 4 3 –1 6.3.6 The Pythagorean identities – 'cos2 + sin2 = 1' 6.3.6A. i) For c = cos , simplify a) ii) 1–c b) c 1 – c2 c) 1 – c2 c2 c) s 1 – s2 For s = sin , simplify a) iii) 2 2 1–s b) 1 – s2 s2 For t = tan , simplify a) 1 + t2 b) t 1 + t2 c) 1 t 1 + t2 Solution All these questions are simply variations on the Pythagorean identities: - 16 - 1 3 –6 n – 6 – cos2 + sin2 1 1 + tan2 sec2 cot2 + 1 cosec2 i) For c = cos we have a) b) c) ii) c 1 – c2 = 1 – cos2 sin2 = sin = cos = 1 – cos2 cos = cot sin 1 – c2 1 – cos2 sin2 = = = tan2 c2 cos2 cos2 For s = sin we have a) 1 – s2 = b) 1 – s2 cos2 cos = = = cosec cot 2 2 s sin sin2 c) iii) 1 – c2 = s 1 – s2 = 1 – sin2 = sin 1 – sin2 = cos2 = cos sin = sec tan cos2 For t = tan we have a) 1 + t2 = 1 + tan2 = sec2 = sec t tan tan = sin cos 2 = 2 = 1+t 1 + tan sec2 1 1 1 c) = = = cos cot 2 2 t 1+t tan 1 + tan tan sec b) 6.3.6B. Eliminate from the equations i) x = a cos y = b sin ii) - 17 - x = a sin , y = b tan Solution x y i) With x = a cos y = b sin we have a = cos and b = sin and so using the Pythagorean identity: x 2 y 2 cos2 + sin2 a + b = 1 ii) With x = a sin , y = b tan we have y = b tan = b sin =b cos sin 1 – sin2 x But sin = a so y=b x/a x2 1 – a bx = a 1– x2 a2 = bx a2 – x2 6.3.6C. For the following values of sin , find the corresponding values of cos and tan without using your calculator, giving your answers in surd form, and assuming that is acute. 2 1 8 7 i) 5 ii) 13 iii) – 17 iv) 25 Solution 2 i) If sin = 5 then cos = 1 – sin2 = So tan = 22 1 – 5 = 4 1 – 25 = 21 21 = 25 5 sin 2/5 2 = = cos 21 21/5 1 ii) If sin = 13 then cos = 1 – sin2 = 1 2 1 – 13 = 1 1 – 169 = - 18 - 168 168 2 42 169 = 13 = 13 So tan = sin 1/13 1 = = cos 42/13 2 42 7 iii) If sin = 25 then cos = 1 – sin2 = 7 2 1 – 25 = So tan = 49 1 – 625 = 576 576 24 = 625 25 = 25 sin 7/25 7 = 24/25 = 24 cos 6.3.6D. If r cos = 3 and r sin = 4 determine the positive value of r, and the principal value of . Solution Square and add to get (r cos )2 + (r sin )2 = r2 cos2 + r2 sin2 = r2 (cos2 + sin2 ) = r2 = 32 + 42 = 9 + 16 = 25 so r = 5 (r is always taken positive) 3 From r cos = 3 we therefore have 5 cos = 3 or cos = 5 from which 3 = cos–1 5 = 53.13 to 2dp 6.3.7 Compound angle formulae 6.3.7A. Prove the following i) sin 3 = 3 sin – 4 sin3 ii) cos 3 = 4cos3– 3 cos iv) cot – tan = 2 cot 2 iii) cos 2 = cos – sin cos + sin - 19 - v) cot2 – 1 cot 2 = 2 cot Solution i) By the compound angle formula we have sin 3 = sin (2 + ) = sin 2 cos + sin cos 2 Using the double angle formula on the sin 2 and cos 2 gives sin 32sin cos cos + sin ( cos2 – sin2) = 3 sin cos2 – sin3 = 3 sin (1 – sin2) – sin3 = 3 sin – 4 sin3 ii) By the compound angle formula we have cos 3 = cos (2 + ) = cos 2 cos – sin sin 2 Using the double angle formula on the sin 2 and cos 2 gives sin 3( cos2 – sin2)cos – sin 2sin cos = cos3 – sin2 cos – 2 (1 – cos2) cos cos – (1 – cos ) cos – 2 (1 – cos2) cos 3 2 = 4cos3– 3 cos iii) cos 2 cos2 – sin2 = cos + sin cos + sin = iv) (cos – sin )(cos + sin ) cos + sin cot – tan = cos sin – sin cos - 20 - = cos – sin cos2 – sin2 cos 2 = =2 sin cos sin 2 on cross mulitplying and using double angle formulae = 2 cot 2 cos2 –1 cos 2 cos2 – sin2 sin2 cot2 – 1 v) cot 2 = = = = sin 2 sin cos cos 2 cot 2 sin 6.3.7B. Without using a calculator or tables evaluate i) sin 15 cos 15 ii) sin 15 v) tan(7/12) vi) cos 75 iii) tan(/12) iv) cos(11/12) Solution In this and the next question you need to be adept at dealing with surds, so if you are a bit rusty on this, go back to Section 1.2.7 for a bit of revision. i) 1 1 1 sin 15 cos 15 = 2 2 sin 15 cos 15 = 2 sin[ 2 (15) ] = 2 sin 30 1 1 1 =2 .2 =4 ii) sin 15 = sin (45 – 30) = sin 45 cos 30 – sin 30 cos 45 1 3 1 1 1 = 2 2 –2 = ( 3 – 1) 2 2 2 = iii) 2( 3 – 1) 4 tan(/12) = tan 15 = tan (45 – 30) = - 21 - tan 45 – tan 30 1 + tan 45 tan 30 1 – 1/ 3 = = 1 + 1/ 3 3–1 ( 3 – 1)2 ( 3)2 – 2 3 + 1 = = 3–1 3 + 1 ( 3 + 1)( 3 – 1) = iv) 4–2 3 =2– 3 2 cos(11/12) = cos 165 = cos (120 + 45) = cos 120 cos 45 – sin 120 sin 45 1 1 3 1 1 2( 3 + 1) =–2 – 2 =– ( 3 + 1) = – 4 2 2 2 2 v) tan 4 + tan3 1+ 3 tan(7/12) = tan 4 + 3 = = 1 – 3 1 – tan 4 tan 3 = (1 + 3)2 1 + 2 3 + 3 4+2 = = – 1–3 –2 2 3 = – (2 + 3 ) vi) cos 75 = cos (30 + 45) = cos 45 cos 30 – sin 30 sin 45 3 1 1 1 = 2 – 2 2 2 = 2( 3 – 1) 4 which of course is just sin (90 – 75) = sin 15 as I'm sure you will have noticed! 6.3.7C. Evaluate i) sin 22.5o ii) o given that cos 45 = 1/ cos 22.5o 2 . iii) Solution This question uses the double angle formulae: sin 2A sin A cos A cos 2A cos2 – sin2 A 2 cos2A – 1 - 22 - tan 22.5o, tan A 1 – 2 sin2A 2 tan A 1 – tan2A i) sin 22.5o = sin (45o/2) = 1 o 2(1 – cos 45 ) by the double angle formula for cos 2A 1 1 1 – = 2 2 = = 2– 2 2 1 o 2(1 + cos 45 ) ii) cos 22.5o = cos (45o/2) = 1 1 21 + 2 = = = sin 22.5o iii) tan 22.5 = = cos 22.5o o = 2( 2 – 1) 4 2– 2 2+ 2 2+ 2 2 2– 2 2+ 2 = 2( 2 + 1) 4 from i) and ii) 2)2 (2 – 6 = 6(2 – 2) 6 6.3.7D. Express the following products as sums or differences of sines and/or cosines of multiple angles i) sin 2x cos 3x ii) sin x sin4x iii) cos 2x sin x iv) cos 4x cos 5x Solution These questions use the compound angle formulae 'backwards'. - 23 - i) We know that sin 2x cos 3x will occur in the expansion of sin (2x + 3x) and sin (2x – 3x). Specifically, we have sin 5x = sin(2x + 3x) = sin 2x cos 3x + sin 3x cos 2x sin (– x) = sin(2x – 3x) = sin 2x cos 3x – sin 3x cos 2x Adding these gives sin 5x + sin (– x) = sin 5x – sin x = 2 sin 2x cos 3x on using the fact that sin x is odd. So 1 sin 2x cos 3x = 2 (sin 5x – sin x) ii) Remembering 'cos (+) = cos cos – sin sin' we know that sin x sin 4x occurs in cos ( x – 4x) = cos 3x, and in cos (x + 4x) = cos 5x. Thus cos 5x = cos (x + 4x) = cos x cos 4x – sin x sin 4x cos 3x = cos (x – 4x) = cos x cos 4x + sin x sin 4x Subtracting and dividing by 2 then gives 1 sin x sin 4x = 2 (cos 3x – cos 5x) iii) We have sin 3x = sin (2x + x) = cos 2x sin x + sin 2x cos x and sin (– x) = – sin x = – sin (2x – x) = – sin 2x cos x + sin x cos 2x from which 1 cos 2x sin x = 2 (sin 3x – sin x) iv) The argument should now be familiar: cos 9x = cos (4x + 5x) = cos 4x cos 5x – sin 4x sin 5x - 24 - cos x = cos (4x – 5x) = cos 4x cos 5x + sin 4x sin 5x and addition gives 1 cos 4x cos 5x = 2 (cos 9x + cos x) 6.3.7E. Prove the following identities (Hint: put P = (A + B)/2, Q = (A – B)/2 in the left-hand sides, expand and simplify and re-express in terms of P and Q) P + Q P – Q i) sin P + sin Q 2sin 2 cos 2 P + Q P – Q ii) sin P – sin Q 2cos 2 sin 2 P + Q P – Q iii) cos P + cos Q 2cos 2 cos 2 P + Q P – Q iv) cos P – cos Q –2sin 2 sin 2 Solution The question hints at the method of proofs of these relations, but there are some shortcuts for the alert. We prove the first in full. i) Put A = P+Q P–Q and B = 2 2 , or P = A + B and Q = A – B. Then sin P + sin Q = sin (A + B) + sin (A – B) = sin A cos B + sin B cos A + sin A cos B – sin B cos A = 2 sin A cos B P + Q P – Q = 2 sin 2 cos 2 ii) - iv) can be proved similarly, or we can use i) along with such ploys as replacing Q by – Q in i) to get ii) and, for example in iii) iii) cos P + cos Q sin (90 – P) + sin (90 – Q) - 25 - 180– (P + Q) Q – P cos = 2 sin 2 2 P + Q P – Q 2 sin 90– 2 cos 2 P + Q P – Q = 2 cos 2 cos 2 6.3.8 Trigonometric equations 6.3.8A. Give the general solution to each of the equations: i) cos = 0 ii) cos = –1 3 iii) cos = – 2 iv) sin = 0 vii) tan = 0 v) sin = –1 viii) tan = –1 vi) sin = 3 ix) tan = 3 Solution These are really examples in inverse functions, and we have in fact done some of them already in RE 6.3.5. We only need to use the results of Section 6.2.5 for the general forms of the inverse trig functions: sin–1x = n + (–1)n PV cos–1 x = 2n PV tan–1 x = n + PV Where PV is the principal value, and n is an integer. i) An equation of the form cos = a has the general solution = 2n PV, so cos = 0 has the general solution (GS) = 2n 2 since the PV of cos–1 0 is 2 . ii) cos = –1 has the GS = 2n = (2n since the PV of cos–1 (– 1) is - 26 - 5 is = –6 = 6 and so the GS is 3 iii) The PV of for cos = – 2 5 = 2n 6 iv) The GS of sin = a is = n – 1n PV. Since the PV of for sin = 0 is = 0, the GS of sin = 0 is just = n v) The GS of sin = –1 is = n (– 1)n 2 since the PV of sin–1 (– 1) is – 2 . vi) No, sin = 3 is not a misprint - just a trap for the unwary! Since, for real angles, , – 1 sin 1 it follows that sin = 3 has no solutions at all. vii) The GS of tan = a is = n + PV and the PV of for tan = 0 is = 0, so the GS of tan = 0 is = n viii) The GS of tan = – 1 is 3 = n + 4 3 since the PV of tan–1 (– 1) is 4 . ix) For tan = 3 the PV of tan–1 ( 3 ) is 3 and so the GS is = n + 3 - 27 - 6.3.8B. Find the general solutions of the equations i) cos 2= 1 iii) cos 2 + sin = 0 v) sec2 = 3 tan – 1 Solution i) ii) iv) sin 2 = sin cos 2 + cos 3 = 0 The GS of cos 2= 1 for 2 is, from the previous question 2 = 2n and so for the GS is = n ii) To solve sin 2 = sin we must get everything in terms of simple sines and cosines, for which we use the double angle formula sin 2 = 2 sin cos = sin This can be rewritten as (2 cos – 1) sin = 0 which shows that 2 cos – 1 = 0 or sin = 0 1 ie either cos = 2 or sin = 0. 1 The GS of sin = 0 is = n The GS of cos = 2 is = 2n 3 which together give the full set of solutions for . An alternative approach to this sort of question is to use the GS of sin A = sin B given in Section 6.2.8, namely A = 2n + B or A = (2n + 1) – B. So the GS of sin 2 = sin will be 2 = 2n + or (2n + 1) – Hence - 28 - = 2n or = (2n + 1) where n is an integer 3 I will leave it as an exercise for you to confirm that this solution is equivalent to the one we obtained above (Evaluate both for n = 0, 1, 2, ...) iii) To solve cos 2 + sin = 0 we must again get everything in terms of simple sines and cosines, for which we use the double angle formula cos 2 = 1 – 2sin2 Then the equation can be rewritten as 1 – 2sin2 sin = 0 or 2 sin2 – sin – 1 = 0 = (2 sin + 1)(sin – 1) 1 Hence either sin = – 2 or sin = 1. 1 1 The GS of sin = 1 is = 2n + 2 . The GS of sin = 2 is = n – (– 1)n 6 . These results can be combined together in the general solution for : 1 2 = 3n + 2 An alternative approach is to use the GS of cos A = cos B given in Section 6.2.8, after converting the sin to a complementary cosine. iv) In cos 2 + cos 3 = 0 use of multiple angle formulae is going to be messy. It is easier to use the result P + Q P – Q cos P + cos Q 2cos 2 cos 2 We have 5 cos 2 + cos 3 = 2 cos 2 cos 2 = 0 - 29 - So 5 cos 2 = 0 or cos 2 = 0 Hence 5 2n + 1 2n + 1 = and therefore = 5 2 2 or 2n + 1 2 = 2 and therefore = (2n + 1) These two apparently different looking results can be combined by noting that when n = 5m + 2 where m is any integer (and of course n can be any integer) 2n + 1 10m + 5 5 = 5 = (2m + 1) So the GS is 2n + 1 = 5 where n is any integer v) sec2 = 1 + tan2 = 3 tan – 1 and so tan2 – 3 tan + 2 = 0 Factorising this quadratic in tan gives (tan – 1)(tan – 2) = 0 So tan = 1 or tan = 2 Therefore = 4 or = 1.11 radians for the PV Hence the GS is = n + 4 , n + 1.11 - 30 - 6.3.9 The a cos + b sin form 6.3.9A. Write the following in the form a) r sin( + ) b) rcos( + ) i) iv) sin – cos 3 cos + 4 sin ii) 3 cos + sin iii) 3 sin – cos Solution i)a) We start by writing sin – cos = r sin( + ) = r sin cos + r cos sin So r cos = 1 and r sin = – 1 Squaring and adding to eliminate gives r = cos = 1 2 2 , and then and sin = – 1 2 The principal value of is then = – 45 = – 4 So we now have sin – cos = 1 1 2 sin – cos = 2 sin – 4 2 2 b) A little more quickly now: sin – cos = r cos( + ) = r cos cos – r sin sin - 31 - So r sin = – 1 and r cos = 1 and again r = 2 , so sin = – 1 2 and cos = – 1 2 The principal value of is then 3 = – 135 = – 4 So 3 sin – cos = 2 cos – 4 ii) a) 3 cos + sin = r sin( + ) = r sin cos + r cos sin So r sin = 3 and r cos = 1 from which r = 2 and then 1 3 cos = 2 and sin = 2 The principal value of is then = 60 = 3 So we have 3 cos + sin = 2 sin + 3 - 32 - b) 3 cos + sin = r cos( + ) = r cos cos – r sin sin So r cos = 3 and r sin = 1 from which r = 2 and then 1 3 sin = – 2 and cos = 2 The principal value of is therefore = – 30 = – 6 So we have 3 cos + sin = 2 cos – 6 iii) a) 3 sin – cos = r sin( + ) = r sin cos + r cos sin So r cos = 3 and r sin = – 1 from which r = 2 and then 1 3 sin = – 2 and cos = 2 The principal value of is then =–6 So we have - 33 - 3 sin – cos = 2 sin – 6 b) 3 sin – cos = r cos( + ) = r cos cos – r sin sin So r sin = – 3 and r cos = – 1 from which r = 2 and then 1 3 cos = – 2 and sin = – 2 The principal value of is therefore 5 = – 150 = – 6 So we have 5 3 sin – cos = 2 cos – 6 iv) a) 3 cos + 4 sin = r sin( + ) = r sin cos + r cos sin So r sin = 3 and r cos = 4 From which r = 5, and then 4 cos = 5 3 and sin = 5 The principal value of is then - 34 - = 36.87 = 0.6435 radians So we now have 3 cos + 4 sin = 5 sin( + 0.64) ) b) 3 cos + 4 sin = r cos( + ) = r cos cos – r sin sin So r sin = – 4 and r cos = 3 and again r = 5, so 3 cos = 5 4 and sin = – 5 The principal value of is then = – 53.13 So 3 cos + 4 cos = 5 cos( – 53.13) 6.3.9B. Determine the maximum and minimum values of each of the expressions in Question A, stating the values where they occur, in the range 0 2. Solution i)a) sin – cos = 2 sin – 4 from A. This has a maximum value of 2 when – 4 = 2 , or 3 = 4 - 35 - It has a minimum value of – 3 2 at – 4 = 2 , or 7 = 4 3 7 So there is a maximum of 2 at = 4 and a minimum value of – 2 at = 4 ii) 3 cos + sin = 2 sin + 3 This has a maximum of 2 at + 3 = 2 , ie = 6 7 3 It has a minimum of – 2 at + 3 = 2 , ie = 6 iii) 3 sin – cos = 2 sin – 6 2 This has a maximum of 2 at – 6 = 2 , ie = 3 5 3 It has a minimum of – 2 at – 6 = 2 , ie = 3 iv) 3 cos + sin = 5 sin( + 0.64) This has a maximum of 5 at + 0.64 = 2 , ie = 0.93 radians 3 It has a minimum of – 5 at + 0.64 = 2 , ie = 4.07 radians 6.3.9C. Find the solutions, in the range 0 2, of the equations obtained by equating the expressions in Question A to a) 1 b) –1. Solution - 36 - i) a) sin – cos = 2 sin – 4 = 1 from A. So 1 sin – 4 = 2 In the range 0 2 the solutions are 3 –4 =4, 4 and therefore 3 =2, 2 b) sin – cos = 2 sin – 4 = – 1, gives 1 sin – 4 = – 2 In the range 0 2 the solutions are 5 –4 =– 4, 4 and therefore = 0, ii) a) 3 cos + sin = 2 sin + 3 = 1 So 1 sin + 3 = 2 In the range 0 2 the solutions are 5 17 +3 = 6 , 6 and therefore - 37 - 5 =2, 3 b) 3 cos + sin = 2 sin + 3 = – 1 So 1 sin + 3 = – 2 In the range 0 2 the solutions are 7 11 +3 = 6 , 6 and therefore 5 3 = 6 , 2 iii) a) 3 sin – cos = 2 sin – 6 = 1 So 1 sin – 6 = 2 In the range 0 2 the solutions are 5 –6 =6, 6 and therefore =3, b) 3 sin – cos = 2 sin – 6 = – 1 So 1 sin – 6 = – 2 In the range 0 2 the solutions are - 38 - 7 11 –6 = 6 , 6 and therefore 4 = 3 , 2 iv) a) 3 cos + 4 sin = 5 sin( + 0.64) = 1 So 1 sin( + 0.64) = 5 = 0.2 In the range 0 2 the solutions are, to two decimal places + 0.64 = – 0.2 = 2.9 and 2 + 0.2 = 6.48 and therefore = 2.3, 5.84 radians b) 3 cos + 4 sin = 5 sin( + 0.64) = – 1 So 1 sin( + 0.64) = – 5 = – 0.2 In the range 0 2 the solutions are + 0.64 = 3.34 and 6.28 and therefore = 2.7, 5.44 - 39 -