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Basic Trigonometric results and trigonometric ratios of arbitrary angles INTRODUCTION : The word Trigonometry in its primary sense signifies the measurement of triangles. From ancient times the Trigonometry also included the establishment of the relations which subsist between the sides , angles and area of a triangle; but now it has a much wider scope and embraces all manner of geometrical and algebraical investigations carried on through the medium of certain quantities called trigonometrical ratios. In every branch of Higher Mathematics, whether Pure or Applied, a knowledge of Trigonometry is of the greatest value. 1. DEFINITION OF ANGLE : Suppose that the straight line in the figure is capable of revolving about the point O, and suppose that in this way it has passed successively from the position OA to the position occupied by OB, OC, OD, …., then the angle between OA and any position such as OC is measured by the which the line OP has undergone in passing from its initial position OA into its final position OC. Moreover the line OP may make any number of complete revolutions through the original position OA before taking up its final position. In Trigonometry angles are not restricted as Geometry, but may be of any magnitude. The point O is called the origin, and OA the initial line; the revolving line OP is known as generating line or the radius vector. 2. MEASUREMENT OF ANGLES : There are several ways of measuring angles in trigonometry. The student is already familiar with the Sexagesimal Measure in which a right angle is equal to 90. There is an absolute measure of an absolute angle which is called Circular or Radian Measure. In this system one unit of angle (one radian) is the angle subtended at the centre of a circle by an arc whose length is equal to the radius. Since the half circumference of a circle of unit radius is . We must have c = 180 where the index c 180 signifies the circular measure. This is loosely written as = 180. Thus 1c and 10 . 180 The student is advised to remember the following angles in radians. c Deg 30 45 60 90 120 135 150 Rad 6 4 3 2 2 3 3 4 5 6 180 210 225 240 270 300 315 330 360 7 6 5 4 4 3 3 2 5 3 7 4 11 6 2 We will be using Sexagesimal system (angles in degrees) in evaluation of trigonometric ratios initially only after which angles will be in radians unless the contrary is specified. 3. DEFINITIONS OF TRIGONOMETRIC FUNCTIONS For an angle between 0 and 90 we define several trigonometric ratios to describe size or magnitude of the angle. Let the initial position 1 acquires a position OB given that BOA = . to OA, we define of the revolving line be OA and the revolving line Choose a point P on OB and draw PM perpendicular MP OP sin , cos ec OP MP OM OP cos , sec OP OM PM OM tan , cot OM PM We can easily note that (i) the values of the above six ratios do not depend upon the choice of P (If P′ is some other point and M′ is the foot of then OPM and OP′M′ are similar) 1 1 1 sin cos (ii) , tan cos ec ,sec , cot , cot sin cos tan cos sin (iii) sin 2 cos2 1, 1 tan 2 sec2 , 1 + cot2 = cosec2 (follow by Pythogorous theorem) Section –I Trigonometric transformations 1. We are already familiar with transformations like sin(90 ) cos , cos(90 ) sin , tan(90 ) cot etc. Let us now show that if is any arbitrary angle then (i) sin( ) sin (ii) cos( ) cos (iii) sin( 180 ) sin (iv) cos(180 ) cos (v) sin( 90 ) cos , (vi) cos(90 ) sin We will give circle proof for above transformations formulas We will first of all assume that is a positive acute angle i.e. 0 90 or 0 2 and consider the circle with centre O and radius r . Let A' OA is the horizontal diameter of the circle. Let the revolving line make an angle with positive directions of x axis . Draw PM perpendicular to x axis and produce it to a point Q on the circle. Then PM MQ and Q 2 is the point (r cos ,r sin ) . This is symbolically written as r cos( ) r cos , r sin( ) r sin on Cancelling r we get cos( ) cos , sin( ) sin we can also get tan( ) tan , cot( ) cot , sec( ) sec cos ec( ) cos ec from the second formula . For instance tan( ) sin( ) sin tan etc. cos( ) cos Now to prove (iii) and (iv) we again consider the same circle (see figure) produce the radius PO in the opposite direction to reach the point Q on the circle Note that OAQ 180 then OM ' OM , M ' Q MP . But co-ordinate sign convection wise Q will have co-ordinates (r cos ,r sin ) . This is symbolically written as r cos(180 ) r cos r sin(180 ) sin . Thus cos(180 ) cos Also tan(180 ) sin( 180 ) sin tan . cos(180 ) cos From these relations we at once get sin( 180 ) sin( 180) sin( ) sin , cos(180 ) cos( 180) cos( ) cos . Also tan(180 ) tan To prove (v) and (vi) we again consider the circle let AOP , OP r POQ 90 Then OM r cos PM r sin OM ' PM r sin QM ' OM r cos Following sign conventions of co-ordinate geometry co-ordinates of Q are (r sin , r cos ) giving us cos(90 ) sin , sin( 90 ) cos 3 Using above six transformations formula’s we can derive all types of transformations formula’s. For example (270 ) sin( 180 90 ) sin( 90 ) cos cos( 270 ) cos(180 90 ) cos(90 ) sin tan( 270 ) cot We will now show that all above results are true for arbitrary i.e. may be any angle. If lies between 90 and 180 then 90 Now sin( 90 ) sin( 90 90 ) = cos(90 ) sin sin( 90) sin( 90 ) cos if lies between 180 and 270 then 180 sin( 90 ) sin( 90 180 ) sin( 270 ) cos cos( 180) cos Thus sin( 90 ) cos in all cases geometrically it is easily seen as sin( 90) cos We can show the validity of transformation formulas for an arbitrary angle θ directly also We will first prove it for 90 θ 180 . If 90 θ 180 then the point P cos θ,sin θ lies in second quadrant (see figure and note that cos θ 0,sin θ 0 ) If P moves in counter clockwise direction for an angle 180 then P reaches a point P in the fourth quadrant. P has co-ordinates cos θ , sin θ . Note that cos θ 0, sin θ 0 cos θ 180 cos θ sin θ 180 sin θ as before 4 Again if P lies in third quadrant then P is cos θ ,sin θ while P is cos θ , sin θ in first quadrant Finally if P lies in fourth quadrant then P is cos θ , sin θ In all the cases we have cos θ 180 cos θ , sin θ 180 sin θ We can similarly show this fact for other transformation formulas. If an angle φ is more than 360 then φ 360n θ where 0 θ 360 and n is a positive integer. Now T-ratio of φ will be same as corresponding T-ratios of θ . For example sin 180 φ sin 180 360n θ sin 180 θ sin θ Thus sin 180 θ sin θ for arbitrary θ The other transformation formulas are also true for all θ R . See more illustrations in 2(d) 2. How to remember all transformation formulas easily ?: Though we have proved all transformation formulas but have not discussed the way to remember all the formulas. Let us discuss four important points in detail. IInd IIIrd 90 θ θ 180 θ 90 θ first 270 θ 180 θ 360 θ 270 θ IVth θ (a). Quadrant determination by assuming that θ is an acute angle. This is shown in the figure. As an example if an angle is of the type 180 θ or 270 θ then it belongs to IIIrd quadrant. (b). Sign of Trigonometric functions : It is necessary to know the sign of signs of sines, cosines and tangents only in various quadrants. The signs of cosec, sec and cot wil be same as that of sin, cos and tan respectively. sin + All + tan + cos + These signs convention is based up on co-ordinate geometry sign convention for above. In the first quadrant all trigonometric functions are positive since x and y co-ordinates are both positive. In the second sin θ is positive since y co-ordinate is positive. In the third tan θ is positive since both x and y are negative and tan θ sin θ . In the fourth cossinθ is positive as cos θ x-coordinate is positive. We can remember it by recalling the AFTER SCHOOL TO COLLEGE A of AFTER stands for all , S of school stands for sin θ T of To stands for tan θ , C of College stands for cosθ 5 Thus, sin 210 0, cos 300 0 Since 210 belongs to third quadrant while 300 belongs to fourth. We can also say sin 180 θ 0,cos θ 0 tan 360 θ 0 , sec 270 θ 0,cot 90 θ 0 etc. ( c). The transformation formulas do not change the type of trigonometric function on both side of it if transformation is being applied on an angle of the type θ ,180 θ ,360 θ . But it changes from sin to cos, cos to sin, tan to cot, cot to tan, sec to cosec, cosec to sec If transformation is being applied on an angle of the type 90 θ , 270 θ Example : If we want to simplify sin 180 θ we note that (i) 180 θ belongs to third quadrant (ii) sin 180 θ 0 (iii) No change in T-ratio will occur sin 180 θ sin θ Again cos 270 θ sin θ . Since cos will change to sin and 270 θ belongs to fourth quadrant where cos is positive. Further note cot θ cot θ . Thus we can remember all formulas easily. We have the following long list of formulas. SET I SET II sin θ sin θ sin 180 θ sin θ cos θ cos θ cos 180 θ cos θ tan θ tanθ tan 180 θ tan θ cot θ cot θ cot 180 θ cot θ sec θ sec θ sec 180 θ sec θ co sec θ co sec θ co sec 180 θ co sec θ SET III SET IV sin 180 θ sin θ sin 360 θ sin θ cos 180 θ cos θ cos 360 θ cos θ tan 180 θ tan θ tan 360 θ tan θ 6 cot 180 θ cot θ cot 360 θ cot θ sec 180 θ sec θ sec 360 θ sec θ cosec 180 θ co sec θ cos ec 360 θ cos ecθ SET V SET VI SET VII sin 90 θ cos θ sin 270 θ cos θ sin 270 θ cos θ cos 90 θ sin θ cos 270 θ sin θ cos 270 θ sin θ tan 90 θ cot θ tan 270 θ cot θ tan 270 θ cot θ cot 90 θ tan θ cot 270 θ tan θ cot 270 θ tan θ sec 90 θ cos ecθ sec 270 θ cos ecθ sec 270 θ cos ecθ co sec 90 θ sec θ co sec 270 θ sec θ co sec 270 θ sec θ TRIGONOMETRIC RATIOS OF ANY ARBITRARY ANGLE: If the angle is greater than 360, then the T-ratios can be found by using the fact that, sin(360 n + ) = sin ; cos(360 n + ) = cos ; tan(360 n + ) = tan , where “n” is an integer. The following steps may be used :Step I: Remove negative sign by using transformations (i) Step II: Divide the angle by 360 and consider the T-ratio of the remainder. Step III : Express the angle as 180 or 360 - , if it is not acute (use of 90 , 270 may be avoided since in 180 , 360 and - cases, the T-ratio remains same). We will illustrate it by examples (i) sin 900 θ sin 2 360 180 θ sin 180 θ sin θ cos 1360 θ cos 3 360 270 θ cos 270 θ sin θ (ii) sin 840 = sin(360 2 + 120) = sin 120 = sin(180 – 60) = sin 60 = (iii) tan(-1200) = -tan 1200 = -tan(3 360 + 120) = -tan 120 = -tan(180 – 60) = -(-tan 60) = 3 3 2 (iv) tan 7π θ tan 6π π θ tan π θ tan θ (v) π 2016π π tan 2015 θ tan θ (Bringing a multiple of 4 near 2015) 2 2 7 π π tan 1008π θ tan θ ( 1008π are 504 resolutions of line) 2 2 π π tan θ tan θ cot θ 2 2 Section –II Given one T-ratio with location of angle how to find other T-ratios In earlier classes we have done that, given sin θ 3/ 5. Find cos θ , tan θ , cot θ ,sec θ etc. We used to solve this problem quickly by constructing a triangle only. And used to determine other T-ratios easily. This used to be a simple and routine problem in earlier classes but at an advanced level it is to be supplemented by more concepts and rigour. For instance if 3 sin θ and 90 θ 180 then as cos θ 1 sin 2 θ 5 We have cos θ 1 sin 2 θ cos θ 1 sin 2 θ 1 cos θ 0 cos θ cos θ sin θ 3/ 5 9 4 Also tan θ cos θ 4 / 5 25 5 TRIGONOMETRIC RATIOS OF 0, 90 AND SOME POSITIVE ACUTE ANGLES : 0 30 45 60 1 1 3 sin 0 2 2 2 1 1 3 cos 1 2 2 2 1 tan 0 1 3 3 Trigonometric ratios of angles which are multiples of 90 1 0 not defined : If n is any integer then the angles 2 n are of great importance in mathematic since sine and cosine become 1 or 0 at these 2 angles. (i) sin n = 0 (ii) cos n = (1)n (iii) tan n = 0 (iv) sin(4n + 1) /2 = 1 (v) sin (4n – 1) /2 = -1 (vi) cos(2n + 1) /2 = 0 (vii) tan (2n + 1) /2 is not defined. 8 Domain, Range of trigonometric functions: T – Function Domain Range sin x R [-1, 1] cos x R [-1, 1] tan x R – {x : x = (2n + 1)/2} R cot x R – {x : x = n} sec x R – {x : x = (2n + 1)/2} (-, -1] [1, ) cosec x R – {x : x = n } R (-, -1] [1, ) SOLVED EXAMPLES 1. Show that SOLUTION:- sin( 90 ) cos(180 ) sec(180 ) 1 tan( 270 ) sin( 360 ) , cos(180 ) cos sec(180 ) sec , tan( 270 ) cot sin( 90 ) cos sin( 360 ) sin Thus LHS of above expression 2. Show that (i) cos ( cos )( sec ) 1 cot ( sin ) sin( 2015 ) sin( 35 ) (ii) sin( 2015 ) sin( 1800 215 ) SOL:- (i) sin( 215 ) sin( 180 35 ) (ii) sin( 600 ) tan( 60 ) cos(1680 ) 600 360 240 LHS sin( 35 ) 1680 360 4 240 sin( 240 ) tan( 240 ) cos( 240 ) tan(180 60 ) tan( 60 ) 3. 3 cos sin( 2 ) 2 Show that 3 cos ec sec(8 ) 2 sin 2 cos 2 9 SOL:- 3 cos sin 2 sin( 2 ) sin cos ec sec 2 sec(8 ) sec( ) sec LHS sin ( sin ) ( sec ) sec sin 2 cos 2 cos 2 4. Show that sin 2015 SOL:- The integer just less than 2015 which is divisible by 4 is 2012 . We write sin 2015 2012 3 sin 2 3 sin 1006 2 2 3 sin cos 2 EXERCISE 1. 2. 3. 4. 5. 6. 7. 8. 9. Find the radian measures corresponding to the following degree measures 15 240 (4) 530 (1) (2) (3) 3730 Find the degree measures corresponding to the following radian measures 3 5 7 (1) (2) -4 (3) (4) 4 3 3 A wheel makes 360 revolutions in one minute .Through how many radians does it turn in one second? Find the degree measure of the angle subtended at the centre of a circle diameter 200 cm by an are of length 22cm. In a circle of diameter 40cm , the length of chord is 20cm. Find the length of minor arc of the chord. If, in two circles, arcs of the same length subtend angles of 60 and 75 at the centre , find the ratio of their radii: Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length: (1) 10cm (2) 15cm (3) 21cm Find the values of the other five trigonometric functions in each of the following problems: 1 3 (1) (2) sin , lies in second quadrant. cos , lies in third quadrant 2 5 4 13 (3) (4) tan , lies in third quadrant. sec , lies in fourth quadrant. 3 5 show that (1) sin 3060 0 (2) cos570 10 1 2 (3) sin 1845 1 2 (4) (7) (10) (13) (16) 10. 1 2 1 sin(330) 2 1 sin 765 = (11) 2 1 cos 855 2 cos 1755 sin 4530 1 2 (5) cos ec(1200) (8) sin 510 2 3 1 2 15 cot 1 4 (14) cos(1125) (17) tan (6) tan 315 1 (9) cos ec(1410) 2 1 11 6 3 (12) tan 1 2 (15) tan(585) 1 (18) sin 13 . 3 3 3 5 3 2 Prove that : 1 2 (1) sin 780 sin120 cos 240 sin 390 (2) sin 300 cos 210 cos 300 sin 210 1 (3) tan 720 cos 270 sin150 cos120 (4) (5) (6) 1 4 sin 600 cos390 cos 480 sin150 1 tan(225)cot (405) tan (765)cot 675 0 sin 8 23 13 35 1 cos cos sin 3 6 3 6 2 (7) cos 24 cos55 cos125 cos 204 cos300 (8) tan 225 cot 405 tan 765 cot 675 0 (9) cos570 sin510 sin (330 cos(390) 0. (10) 2sin 2 (12) 2sin 2 (14) tan (15) 3 2cos2 sec2 6 4 4 3 6 cos ec 7 cos2 0 6 3 1 2 (11) sin 2 (13) cot 2 6 6 11 2 3 17 3 4 3 2sin cos ec 2 4cos 2 . 3 3 4 4 6 2 cos( x)cos( x) cot 2 x sin( x)cos x 2 11 cos2 3 cos ec tan 2 4 1 2 5 3tan 2 6 6 6 11. (16) sin 180 A cot 90 A cos 360 A sin A tan 180 A tan 90 A sin A (17) sin(180 )cos(90 ) tan(270 )cot (360 ) 1 sin(360 )cos(360 )cos ec( )sin(270 ) (18) cos ec(90 ) cot (450 ) tan(180 ) sec(180 ) 2 cos ec(90 ) tan (180 ) tan (360 ) sec( ) (19) 3 3 sin cos cot sin sin cot 2 2 2 2 2 cos(2 )cos ec(2 ) tan 2 1. (20). sec cos cot 2 3 3 (21). cos x cos(2 x) cot x cot(2 x) 1 2 2 In a ABC, prove that: C A B (1) cos(A + B) + cos C = 0 (2) cos sin 2 2 A B C (3) tan cot 2 2 Answers 1. 2. (1) (1) 3. 12 7. (1) 8. (1) (2) (3) (4) (5) (10) (15) /12 42 57 16 2/15 (2) (2) - 5/24 -229 5′ 27 4. 12 36 5. (2) 1/5 (3) (3) 4/3 (3) 300 20 cm 3 7/25 (4) (4) 53/18 420 6. 5:4 3 2 1 ,cos ec ,sec 2, tan 3,cot 2 3 3 5 4 5 3 4 cos ec ,cos ,sec , tan ,cot 3 5 4 4 3 4 5 3 5 3 sin ,cos ec ,cos ,sec ,cot 5 4 5 3 4 12 13 5 12 5 sin ,cos ec ,cos , tan ,cot 13 12 13 5 12 1 1 3 (6) 0 (7) (8) 2 2 3 2 ½ (11) -½ (12) ½ (13) 3 1 1 1 (16) (17) (18) 2 2 2 sin 12 (9) -1 (14) -1 -½ Section –III TRIGONOMETRIC RATIOS OF COMPOUND ANGLES When we learn a new function, we come to know about various peculiar properties of these functions. Trigonometric functions in general do not satisfy the relations sin A B sin A sin B tan A B tan A tan B The actual and mathematically correct expansion of sin A B, tan A B initially look very astonishing but later we realize that almost entire trigonometry run around them, one single formula sin A B sin A cos B cos Asin B (*) is capable of generating almost all formulas of trigonometry. If we assume the validity of (*) (Proof later) then sin A B sin A cos B cos Asin B sin A cos B cos Asin B cos A B sin A B sin A cos B cos A sin B 2 2 2 cos A cos B sin Asin B cos A B cos A cos B sin Asin B cos A cos B sin Asin B sin A cos B cos A sin B sin A sin B cos A sin B tan A tan B sin A B cos A cos B cos A sin B tan A B cos A cos B sin A sin B 1 tan A tan B cos A B cos A cos B sin A sin B cos A cos B cos A cos B tan A B tan A tan B 1 tan A tan B cot A B cot A B cot A cot B 1 1 1 tan A tan B cot B cot A tan A B tan A tan B cot A cot B 1 cot A cot B 1 cot B cot A cot B cot A In the next two chapters we will observe that by adding, subtracting two A B formulas of same kind we can express sums as products and products as sums. We will further observe that main trigonometric formulas sin 2 A, cos 2 A, sin 3 A are also derived from these A B formulas only. In this chapter we will be discussing questions on following six formulas :- sin A B sin A cos B cos Asin B 13 sin A B sin A cos B cos A sin B cos A B cos A cos B sin Asin B cos A B cos A cos B sin Asin B tan A B tan A tan B tan A tan B , tan A B 1 tan A tan B 1 tan A tan B In addition to these there are two ‘surprise formulas’ sin A B . sin A B sin 2 A sin 2 B and cos A B . cos A B cos 2 A sin 2 B which are easily proved by expanding and multiplying. We call them surprise formulas because the pattern is false for other trigonometric functions. For example tan A B . tan A B tan 2 A tan 2 B . We will now give formal proof of these formulas To prove the formulae sin A B sin A cos B cos A sin B, cos A B cos A cos B sin A sin B , see figure Let XOX ' A, and X ' OX '' B; then XOX '' A B. On OX '' , the final position of initial line of the compound angle A B , take any point P , and draw PQ and PR perpendicular to OX and OX ' respectively; also draw RS and RT perpendicular to OX and PQ respectively. Now, Since PQ RS PT RS PT OP OP OP OP RS OR PT PR . . sin A.cos B cos A.sin B OR OP PR OP TPR 90 TRP TRO ROS A; sin A B Thus sin A B sin A cos B cos A sin B 14 Also cos A B OQ OS TR OS TR OP OP OP OP OS OR TR PR . . OR OP PR OP cos A.cos B sin A.sin B Thus cos A B cos A cos B sin A sin B Finally tan A B But RS PT RS PT OS OS OS OS TR TR TP 1 1 . OS TP OS PQ RS PT OQ OS TR RS TR tan A, and tan A; OS TP TP PR tan B OS OR tan A tan B Thus tan A B 1 tan A tan B also the triangles ROS and TPR are similar We can similarly derive tan A B tan A tan B 1 tan A tan B To prove the formulas sin A B sin A cos B cos A sin B cos A B cos A cos B sin A sin B Let XOX ' A, and X ' OX '' B; then XOX '' A B. On OX ' take any point P , and draw PQ and PR perpendicular to OL and OM respectively. Draw RS and RT perpendicular to OL and QP respectively. Now, PQ RS PT RS PT OP OP OP OP RS OR PT PR . . OR OP PR OP sin A B 15 sin A.cos B cos A.sin B Since TPR 90 TRP X '' RT X ' OX A; sin A B sin A cos B cos A sin B. Also OQ OS RT OS RT OP OP OP OP OS OR RT RP . . OR OP RP OP cos A B cos A.cos B sin A.sin B SOLVED EXAMPLES 1. cos sin sin = sin( ) 4 4 4 4 Show that cos Sol : If we examine carefully LHS cos A cos B sin A sin B where A 4 , B LHS cos( A B) cos cos sin 4 4 2 2. cos 2 33 cos 2 75 2 2 69 2 21 sin sin 2 2 Show that Sol : cos 2 33 cos 2 57 1 sin 2 33 1 sin 2 57 sin 2 57 sin 2 33 sin 57 33sin 57 33 sin 90 sin 24 sin 24 69 21 69 21 69 21 sin sin 2 = sin . sin sin 45 sin 24 2 2 2 2 2 2 2 LHS 3. Show that RHS = sin 24 sin 45 sin 24 2 cos65 sin 65 cot 70 cos65 sin 65 1 tan 65 1 tan 65 tan 45 tan 65 1 tan 45 tan 65 ( on dividing by cos 65 ) = tan 45 65 16 tan( 20) = tan 20 cot 70 4 4. Show that in a triangle whose angles are A, B, C tan A tan B tan C tan A.tan B.tan C Sol : A B 180 C tan( A B) tan(180 C ) tan A tan B tan C 1 tan A tan B 1 tan A tan B tan C tan A tan B tan C tan A tan B tan C tan A tan B tan C NOTE: In general , If ' tan' of both sides is taken we get a result of the type Sum of tangents = Product of tangents For example tan 10 tan 7 5. tan 10 tan 7 tan 1 tan 7 tan tan 10 tan 7 tan tan 10 tan 7 tan If and are the solutions of the equation a tan b sec c, then show that tan Sol : 2ac a c2 2 a tan b sec c c a tan 2 b 2 sec 2 On arranging c a tan b sec (*) c 2 a 2 tan 2 2ac tan b 2 (1 tan 2 ) tan 2 (a 2 b 2 ) 2ac tan (c 2 b 2 ) 0 (**) Since and are the solutions of (*) . Therefore , tan and tan are roots of equation (**). tan tan 2ac c2 b2 and tan tan a 2 b2 a 2 b2 2ac 2ac tan tan (a b 2 ) = = 2 tan( ) 2 2 c b a c2 1 tan tan 1 2 a b2 2 EXERCISE 17 1. 2. 3. 24 3 3 3 , cos where and 2 . 25 5 2 2 3 4 Show that (i) sin (ii) cos 5 5 3 12 3 16 sin ,0 ,cos , show that tan 5 2 13 2 63 If cos Prove the following identities: (1) cos A B cos B sin A B sin B cos A (2) sin 3A cos A – cos 3A sin A = sin 2A (3) cos(30 + A) cos(30 - A) – sin(30 + A)sin(30 - A) = ½ (4) sin(60 - A) cos(30 + A) + cos(60 - A)sin(30 + A) = 1 (5) sin 2 cos + cos 2 sin = sin 4 cos - cos 4 sin (6) cos 2 cos + sin 2 sin = cos (7) cos 4 cos + sin 4 sin = cos 2 cos - sin 2 sin (8) cos 4 cos - sin 4 sin = cos 3 cos 2 - sin 3 sin 2 (9) tan tan 1 tan tan tan (10) tan 4 A tan 3 A tan A 1 tan 4 A tan 3 A (12) sin(150 + x) + sin(150 - x) = cos x (15) cos2 2x – cos2 6x = sin 4x sin 8x (11) cot( ) cot 1 cot cot cot( ) (13) 3 3 cos x cos x 2 sin x 4 4 (14) cos x cos x 2 sin x 4 4 (16) tan 3x tan 2x tan x = tan 3x – tan 2x – tan x (17) sin(60 + A) – sin(60 – A) = sin A (18) cos 30 A cos 30 A 3 cos A (19) cos cos 2 sin 4 4 (20) sin2 5A – sin2 2A = sin 7A sin 3A cos 2A cos 5A = cos2 (22) (23) sin sin sin sin B C cos B cos C sin sin sin (21) sin sin sin sin C A cos C cos A 0 sin A B cos A cos B 18 0 7A 3A sin 2 2 2 (24) tan( - ) + tan( - ) + tan( - ) = tan( - )tan( - )tan( - ) (25) cos2 + cos2 - 2cos cos cos( + ) = sin2 ( + ) (26) sin2 + sin2 + 2sin sin cos( + ) = sin2( + ) (27) cos ec (29) cos x cos y sin x sin y sin( x y) 4 4 4 4 (30) tan x 2 4 1 tan x 1 tan x tan x 4 (31) sin(n 1) x sin(n 2) x cos(n 1) x cos(n 2) x cos x (32) 3 3 cos x cos x 2 sin x 4 4 (33) sin 2 6 x sin 2 4 x sin 2 x sin10 x (34) (35) cot x cot 2 x cot 2 x cot 3x cot 3x cot x 1 (36) cos2 cos2 sin 2 ( ) sin 2 ( ) cos2( ) cos ec cos ec cot cot (28) cos 1 tan tan sec sec cos2 2 x cos2 6 x sin 4 x sin8 x 4. If 2 tan cot tan , prove that cot 2 tan( ) 5. If tan 6. If A B 7. If A + B = 225, prove that 8. Prove that : 9. If sin( ) 1 and sin( ) n sin cos , show that tan( ) (1 n) tan 1 n sin 2 4 , prove that : (i) (1 tan A)(1 tan B) 2 (ii) (cot A 1)(cot B 1) 2 cot A cot B 1 1 cot A 1 cot B 2 tan 2 2 x tan 2 x tan 3x tan x 1 tan 2 2 x tan 2 x 1 , where 0 , , then find the values of 2 2 tan( 2 ) and tan( 2 ) . 19 10. Prove that 1 cot( x a) cot( x b) sin( x a) sin( x b) sin( a b) 11. If tan m 1 , tan , show that + = /4. m 1 2m 1 Section –IV Expressing product as sum and sum as products Product as sum : In the last section we have learnt four important addition and subtraction formulas. If we add or subtract two such formulas of same kind we get formulas expressing product as a sum and sum as a product. Indeed by adding sin A B and sin A B formulas we easily get sin A B sin A B 2 sin Acos B . If we write it in the reverse order we get a formula expressing the product of the type 2 sin cos as a sum of two sines. For example, 2 sin 65 cos 25 sin 65 25 sin 65 25 sin 90 sin 30 1 1/ 2 3 / 2 . Similarly, we can have formulas 2 cos Asin B sin A B sin A B 2 cos Acos B cos A B cos A B 2 sin Asin B cos A B cos A B One has to be careful in applying last formula in which comes at first place Sum as product : - If in all above four formulas we put A B C , A B D then by writing in the CD CD cos reverse direction we easily get sin C sin D 2 sin 2 2 CD CD sin C sin D 2 cos sin 2 2 CD CD cos C cos D 2 cos cos 2 2 CD CD cos C cos D 2 sin sin 2 2 Once again we have to be careful in applying formula number 4. The intricacies of these formulas will be more clear when we do sufficient number of problems. For instance the formula 2 cos Asin B sin A B sin A B may be avoided if we write 2 cos Asin B 2 sin B cos A sin B A sin B A sin A B sin A B A will never appear in cos C cos D because of cos cos . Let us illustrate usage of these formulas by means of examples. SOLVED EXAMPLES 1. 1 Show that sin140 cos 40 cos10 2 Sol : sin 140 cos 40 1 2 sin 140 cos 40 2 1 sin( 140 40) sin( 140 40) 2 20 1 sin 180 sin 100 1 sin100 1 cos10 2 2 2 1 Show that sin 10 sin 50 sin 70 8 1 1 LHS sin 102 sin 50 sin 70 sin 10cos (50 70) cos(50 70) 2 2 ( on applying 2 sin A sin B cos( A B) cos( A B)] 1 = sin 10 cos( 20 ) cos120 2 1 1 1 sin 10cos 20 ( cos( 20) cos 20 and cos 120 ) 2 2 2 1 1 1 1 sin 10 cos 20 sin 10 2 sin 10 cos 20 sin 10 2 4 4 4 1 1 sin (10 20) sin( 10 20) sin 10 4 4 1 1 sin 30 sin 10 sin 10 1 sin 30 1 sin 10 1 sin 10 1 = 4 4 4 4 4 8 2. Sol : 3. 1 Show that sin A sin(60 A)sin(60 A) sin 3 A 4 Sol : LHS 4. 1 1 1 1 sin A cos 2 A sin A cos 2 A sin 3 A 2 2 2 4 1 1 1 sin 3 A sin A sin A sin 3 A 4 4 4 Show that sin( B C ) cos( A D) sin( C A) cos( B D) sin( A B) cos(C D) 0 1 sin A 2 sin( 60 A) sin( 60 A) 2 1 sin A cos( 2 A) cos120 2 1 [ sin( B C A D) sin( B C A D) sin( C A B D) sin( C A B D) 2 sin( A B C D) sin( A B C D)] = 0 ( sin( B C A D) gets cut with sin( C A B D) etc ) Show that cos 4 cos10 2 cos 7 cos 3 CD CD cos By applying cos C cos D 2 cos 2 2 4 10 4 10 cos 4 cos10 2 cos cos 2 cos 7 cos 3 2 2 Show that cos18 sin 18 2 cos 63 LHS cos18 sin 18 sin 72 sin 18 72 18 72 18 2 cos sin 2 cos 45 sin 27 2 cos 63 2 2 LHS 5. Sol: 6. Sol: 21 7. Sol: 8. Sol: sin 5 x sin 3 x tan 4 x cos 5 x cos 3x 5 x 3x 5 x 3x 2 sin cos 2 sin 4 x cos x 2 2 tan 4 x = LHS 5 x 3x 5 x 3x 2 cos 4 x cos x 2 cos cos 2 2 Show that Show that sin x sin 3x sin 5x sin 7 x 4 cos x cos 2 x cos 4 x We will use sin C sin D formula twice on correct pairs. The correct pairs are sin x , sin 7 x ; sin 3 x and sin 5x LHS sin x sin 7 x sin 3x sin 5 x 2 sin 4 x cos 3x 2 sin 4 x cos x 2sin 4 x cos3x cos x 2 sin 4 x.2 cos 2 x cos x = 4 cos x cos 2x sin 4x 9. Show that Sol: LHS sin 8 A cos 5 A cos12 A sin 9 A cot 4 A cos 8 A cos 5 A cos12 A cos 9 A sin 13 A sin 3 A sin 21A sin 3 A 2sin8 A cos5 A 2cos12 A sin 9 A = 2cos8 A cos5 A (2cos12 A cos9 A) cos13 A cos 3 A (cos 21A cos 3 A) sin 13 A sin 21A 2 sin 17 A cos 4 A CD CD cos ( Applying sin C sin D 2 sin cos13 A cos 21A 2 sin 17 A sin 4 A 2 2 CD CD sin and cos C cos D 2 sin ) cot 4 A 2 2 tan( A B) k 1 If sin 2 A k sin 2 B prove that . tan( A B) k 1 sin 2 A k sin 2 A sin 2 B k 1 We (applying componendo & divenendo ) sin 2 B 1 sin 2 A sin 2 B k 1 10. Sol: 2 A 2B 2 A 2B cos sin( A B) cos( A B) k 1 2 2 = 2 A 2B 2 A 2B cos( A B) sin( A B) k 1 2 cos sin 2 2 2 sin tan( A B) k 1 . tan( A B) k 1 EXERCISE 5 1 sin 12 12 2 1. Show 2 sin 2. Show that sin 25 cos115 1 (sin 40 1) 2 22 1 8 3. Show that cos 40 cos 80 cos160 4. Show that cos 5. 4 cos12 cos 48 cos 72 cos 36 6 cos 3A sin 2A – cos 4A sin A = cos 2A sin A 7. Prove the following identities: 5 1 cos 12 12 4 (1) sin 2x + 2 sin 4x + sin 6x = 4 cos2 x sin 4x (2) sin + sin 3 + sin 5 + sin 7 = 4cos cos 2 sin 4 (3) cos 7x + cos 5x + cos 3x + cos x = 4 cos x cos 2x cos 4x (4) cot 4x(sin 5x + sin 3x) = cot x(sin 5x – sin 3x) (5) cos9 x cos5 x sin 2 x sin17 x sin 3x cos10 x (6) sin 5 x sin 3x tan 4 x cos5 x cos3x (7) sin x sin y x y tan cos x cos y 2 (8) sin x sin 3x tan 2 x cos x cos3x (9) sin x sin y x y tan cos x cos y 2 (10) tan 5 tan 3 4cos 2 cos 4 tan 5 tan 3 (11) sin x sin 3x 2sin x sin 2 x cos 2 x (12) sin 5x 2sin 3x sin x tan x cos5 x cos x (13) (sin 7 x sin 5 x) (sin 9 x sin 3x) tan 6 x (cos7 x cos5x) (cos9 x cos3x) (14) cos 4 x cos3x cos 2 x cot 3x sin 4 x sin 3x sin 2 x (15) sin 3x + sin 2x – sin x = 4 sin x cos (16) cos cos3 tan 2 sin 3 sin (17) sin 2 sin 3 cot cos 2 cos3 2 (18) cos 4 cos 5 tan sin sin 4 2 (19) cos 2 cos12 tan 5 sin12 sin 2 (20) (22) cos 2 3 cos3 sin 2 3 sin 3 sin sin 4 cos cos 4 x cos 2 cot tan (21) 3 2 23 cos 3 cos 3 sin 3 sin 3 tan 2 8. (23) cos 3A + sin 2A – sin 4A = cos 3A(1 – 2sin A) (24) sin 3 - sin - sin 5 = sin 3(1 – 2cos 2) (25) cos + cos 2 + cos 5 = cos 2(1 + 2cos 3) (26) sin - sin 2 + sin 3 = 4sin /2 cos cos 3/2 (27) sin 3 + sin 7 + sin 10 = 4sin 5 cos 7/2 cos 3/2 (28) sin A + 2sin 3A + sin 5A = 4sin 3A cos2 A (29) sin 2 sin 5 sin tan 2 cos 2 cos5 cos (31) cos7 cos3 cos5 cos cot 2 sin 7 sin 3 sin 5 sin (32) sin( + + ) + sin( - - ) + sin( + - ) + sin( - + ) = 4sin cos cos (33) cos( + - ) – cos( + - ) + cos( + - ) – cos( + + ) = 4sin cos sin (34) sin 2 + sin 2 + sin 2 - sin 2( + + ) = 4sin ( + )sin( + )sin( + ) (35) cos + cos + cos + cos( + + ) = 4cos (36) cos( + )sin( - ) + cos( + )sin( - ) + cos( + )sin( - ) + cos( + )sin( - ) = 0 (37) sin cos( + ) – sin cos( + ) = cos sin( - ) (38) cos cos( + ) – cos cos( + ) = sin sin( - ) (39) sin( - ) + sin( - ) + sin( - ) + 4sin (40) sin (sin 3 + sin 5 + sin 7 + sin 9) = sin 6 sin 4 (41) sin sin 3 sin 5 sin 7 tan 4 cos cos3 cos5 cos7 (42) sin x sin3x sin5x sin 7 x 4cos x cos 2 x sin 4 x (43) 2 cos (44) tan 60 tan 60 (45) cos 20 cos100 cos100 cos140 cos140 cos 200 (46) sin 2 13 cos .sin (30) sin sin 2 sin 4 sin 5 tan 3 cos cos 2 cos 4 cos5 2 2 sin cos 2 2 sin cos 2 2 0 9 3 5 cos cos 0 13 13 13 2 cos 2 1 2 cos 2 1 3 4 7 3 11 sin sin sin 2 sin 5 2 2 2 If cos( A B).sin C D cos( A B) sin( C D) prove that tan A tan B tan C tan D 0 24 9. Show that sin x sin y sin(x – y) + sin y sin z sin(y – z) + sin z sin x sin(z – x) + sin(x – y) sin(y – z) sin(z – x) = 0 for all x, y, z. Section –V Trigonometric Functions of multiple angles Almost all trigonometric formula have been generated by sin( A B ) formula Now we will observe that most useful formulas being generated by these formula’s. The formulas express Trigonometric functions of the angle n A in terms of Trigonometric Functions of angles A where n is an integer ( in general n can be positive rational ) we start deriving these formula’s.We will put frequently used formulas in a block. sin 2 A sin( A B) sin A cos A cos A sin A 2 sin Acos A Thus sin 2 A 2 sin A cos A We can similarly derive, sin 4 A 2 sin 2 A cos 2 A, sin 8 A 2 sin 4 A cos 4 A, sin A 2 sin A A cos 2 2 A A A 2 sin n1 cos n1 n 2 2 2 Again cos 2 A cos ( A A) cos A cos A sin A sin A cos 2 A sin 2 A sin But cos 2 A sin 2 A cos 2 A (1 cos 2 A) 2 cos 2 A 1 1 2 sin 2 A Thus cos 2 A cos 2 A sin 2 A 2 cos 2 A 1 1 2 sin 2 A We can similarly derive cos 4 A cos 2 2 A sin 2 2 A 2 cos 2 2 A 1 1 2 sin 2 2 A A A A A cos A cos 2 sin 2 2 cos 2 1 1 2 sin 2 2 2 2 2 tan A tan A 2 tan A Again tan 2 A tan( A A) . 1 tan A tan A 1 tan 2 A A 2 tan 2 tan A 2 tan 2 A 2 Thus tan 2 A Also tan 4 A or tan A 2 1 tan A 1 tan 2 2 A 2 A 1 tan 2 2 2 tan A 1 tan A we can easily get sin 2 A , cos 2 A 2 1 tan A 1 tan 2 A Or cos 2 A sin 2 A (1 sin 2 A) sin 2 A 3A formulas :- sin 3 A sin(2 A A) sin A cos 2 A cos A sin 2 A = sin A (1 2 sin 2 A) cos A.2 sin A cos A = sin A (1 2 sin 2 A) 2 sin A(1 sin 2 A) = 3sin A 4 sin 3 A cos3 A cos( A 2 A) cos A cos 2 A sin Asin 2 A = cos A (2 cos 2 A 1) 2 sin 2 A cos A = cos A(2 cos 2 A 1) 2 (1 cos 2 A) cos A = 4cos3 A 3cos A tan 3 A tan( A 2 A) tan A tan 2 A 1 tan A tan 2 A 25 2 tan A 3 1 tan 2 A = 3tan A tan A = 2 tan A 1 3tan 2 A 1 tan A. 1 tan 2 A sin 3 A 3sin A 4 sin 3 A tan A Thus cos 3 A 4 cos 3 A 3 cos A tan 3 A 3 tan A tan 3 A 1 3 tan 2 A (i) (ii) The sine, cosine, tangents of angle of n A when n 4 can also be found. We will observe that if n is a positive integer. cos nA will be a polynomial of degree n in cos A for all n . sin nA will be a polynomial of degree n in sin A if n is odd . (iii) tan nA n C1 tan A n C3 tan 3 A n C5 tan 5 A..... 1 n C2 tan 2 A n C4 tan 4 A..... The proof of (i), (ii), (iii) involve three branches of algebra which are complex numbers, binomial theorem and polynomial theory. We will limit our solves upto n 5 . The general proof is beyond the scope of this book. 2 For example : cos 4 A 2 cos 2 2 A 1 2 2 cos 2 A 1 1 8 cos 4 A 8 cos 2 A 1 or cos 4 A Real part of cos 4 A i sin 4 A 4 Real part of cos A i sin A Real part of cos A 4 4 C1 cos A i sin A 4C2 cos A i sin A 4C3 cos A i sin A 4C4 i sin A 3 2 cos 4 A 6 cos 2 A sin 2 A sin 4 A 2 1 3 2 cos 4 A 6 cos 2 A 1 cos 2 A 1 cos 2 A 8 cos 4 A 8 cos 2 A 1 5 sin 5 A Imaginary part of cos 5 A i sin 5 A I.P of cos A i sin A This time writing only Imaginary part we get i sin 5 A 5C1 cos A i sin A 5C3 cos A i sin A 5C5 i sin A 4 1 2 3 5 sin 5 A 5 sin Acos A 10 sin 3 A cos 2 A sin A 4 5 2 5sin A 1 sin 2 A 10sin3 A 1 sin2 A sin5 A 16 sin 5 A 20 sin 3 A 5 sin A (*) The above discussion may lead to something very basic in mathematics that sin x , or cos x are solutions of polynomial equations with integer coefficients. For example If A 18 then 5A 90 from (*) 1 16 x 5 20 x 3 5 x where x sin A 5 3 or 16 x 20 x 5 x 1 0 sin 18 is a solution of a polynomial equation with integer coefficients. Actually numbers which are solutions of polynomial equations with integer coefficients are called algebraic numbers. The number is not algebraic but cos 20 is algebraic since 26 4 cos 60 4 cos3 20 3 cos 20 (by formula cos 3 A 4 cos3 A 3 cos A ) 1 / 2 4 x 3 3x where x cos 20 3 8x 6 x 1 0 cos 20 is algebraic We will later see that sin 18 is a solution of a polynomial equation with much lesser degree. SQUARE, CUBE FROM POWER REMOVAL : From 2 A formula’s we can get rid of power 2,3,4 easily. For example : 1 cos 2 A 1 cos 2 A , sin 2 A 2 2 cos 3 A 3 cos A 3 sin A sin 3 A cos3 A , sin 3 A 4 4 cos 2 A 1 2 cos 2 A cos 2 2 A 4 1 cos 4 A 1 2 cos 2 A 3 1 1 2 cos 2 A cos 4 A 8 2 8 4 2 1 cos 2 A cos 4 A cos 2 A 2 2 (**) 3 1 1 Similarly sin 4 A cos 2 A cos 4 A etc. From these fourth power formula we easily get 8 2 8 cos 4 8 cos 4 3 5 7 3 cos 4 cos 4 by using (**) several times. 8 8 8 2 SOLVED EXAMPLES Example 1 : Prove that cot - cot 2 = cosec 2 Solution : cot cot 2 cos cos 2 sin 2 cos cos 2 sin sin sin 2 sin sin 2 sin 2 sin cos ec 2 RHS sin sin 2 sin sin 2 Example 2 : Prove that 1 + cot 2 cot = cosec 2 cot Solution : 1 cot 2 cot 1 cos 2 cos sin 2 sin sin 2 sin cos 2 cos cos2 cos 1 . sin 2 sin sin 2 sin sin sin 2 cot cos ec2 RHS 27 Example 3 : tan + cot = 2cosec 2 Solution : LHS tan cot 2 2 sin cos sin cos sin 2 cos 2 1 cos sin sin cos sin cos 2 2 cos ec 2 sin 2 2 Example 4 : A A sin cos 1 sin A 2 2 Solution : LHS sin 2 Example 5 : Prove that Solution : LHS A A A A cos 2 2 sin cos 1 sin A 2 2 2 2 cos 2 tan 45 1 sin 2 cos 2 cos 2 sin 2 cos sin cos sin 2 1 sin 2 cos sin 2 cos sin cos sin 1 tan tan 45 tan tan 45 cos sin 1 tan 1 tan 45 tan OR sin 2 2 sin cos cos 2 2 4 4 tan 4 1 sin 2 2 1 cos 2 2 cos 2 4 (Applying sin A 2 sin 1 cos A 2 cos 2 A A cos where role of A is being played by 2 and 2 2 2 A with A as 2 ) 2 2 Example 6 : Prove that cos cos sin sin 4cos2 Solution : On squaring 2 LHS 2 2 cos 2 cos 2 2 cos cos sin 2 sin 2 2 sin sin 2 2cos cos sin sin 2 2 cos 21 cos 2.2 cos 2 4 cos 2 2 2 (Applying 1 cos A 2 cos 2 RHS 28 A with A as ) 2 Example 7 : Prove that cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1 Solution : cos 6x 2 cos 2 3x 1 ( cos 2 A 2 cos 2 A 1 with A as 3x ) 2 2 4cos3 x 3cos x 1 32cos6 x 48cos4 x 18cos2 x 1 Example 8 : A Prove that sin A 1 2sin 2 45 2 Solution : A RHS 1 2 sin 2 45 1 2 sin 2 2 A 45 2 cos 2 cos90 A sin A Example 9 : Prove that cos2 sin 2 sin 2 4 4 Solution : On removing squares 1 cos 2 4 cos 2 4 2 1 cos 2 2 1 sin 2 2 2 1 cos 2 2 1 sin 2 sin 2 2 2 4 On subtracting we get the desired result Example 10 : Prove that Solution : LHS 1 sin 2 cos 2 tan 1 sin 2 cos 2 1 cos 2 sin 2 2sin 2 2sin cos 1 cos 2 sin 2 2cos 2 2sin cos 2 sin sin cos tan RHS 2 cos cos sin Example 11 : Prove that sin 2 x 2sin 4 x sin 6 x 4cos 2 x sin 4 x Solution : LHS sin 2 x sin 6 x 2sin 4 x 2 sin 4x cos 2x 2 sin 4x 2 sin 4 x1 cos 2 x 2 sin 4 x.2 cos 2 x 4 cos 2 x sin 4 x RHS 29 Example 12 : Prove that 2 cos Solution : 1 cos 4 2 cos 2 2 2 2 2 cos 4 2 2 cos 4 4 cos 2 2 On taking square root 2 2 cos 4 2 cos 2 On adding 2 on both sides we get 2 2 2 cos 4 2 2 cos 2 21 cos 2 2.2 cos 2 On taking the square root again we get 2 2 2 cos 4 2 cos as desired Example 13 : Prove that tan A + 2 tan 2A + 4 tan 4A + 8 cot 8A = cot A Solution : LHS tan A 2 tan 2 A 4 tan 4 A 8cot 8 A cot A tan A 2 tan 2 A 4 tan 4 A 8cot 8 A cot A (Subtracting and adding cot A ) Now cot A tan A cos A sin A sin 2 A cos 2 A sin A cos A sin A cos A cos 2 A .2 2 cot 2 A 2 sin A cos A (*) LHS 2cot 2 A 2 tan 2 A 4 tan 4 A 8cot 8 A cot A 4cot 4 A 4 tan 4 A 8cot 8 A cot A (using (*) for A replaced by 2 A ) 8cot 8 A 8cot 8 A cot A (Again using (*)) cot A RHS Example 14 : Prove that Solution : sec 8 A 1 sec 4 A 1 sec8 A 1 tan 8 A sec 4 A 1 tan 2 A 1 1 1 cos 8 A cos 4 A cos 8 A . 1 cos 8 A 1 cos 4 A 1 cos 4 A 2 sin 2 4 A cos 4 A 2 sin 4 A cos 4 A sin 4 A , . 2 sin 2 2 A cos 8 A cos 8 A 2 sin 2 2 A tan 8 A . 2 sin 2 A cos 2 A tan 8 A 2 sin 2 2 A tan 2 A 30 A if tan A 3 / 4 2 Example 15 : Find the values of tan Solution : A 2 2 x where x tan A tan A A 1 x2 2 1 tan 2 2 2 tan Now tan A 3 / 4 A is either in first quadrant (or 2n to 2n quadrant (i.e. from 2n to 2n If A is from 2n to 2n 2 then If A is from 2n to 2n tan 2. 2 ) OR in the third 3 ) 2 A A 1 A tan 0 tan is from n to n 4 2 2 3 2 3 A 3 then is from n to n 2 2 4 2 A A 0 tan 3 2 2 EXERCISE 1. x x x Find sin ,cos and tan in each of the following : 2 2 2 4 1 (1). (2). tan x , x in quadrant II cos x , in quadrant III 3 3 1 (3). tan x , in quadrant II 4 Prove the following identities: (1) cosec 2 - cot 2 = tan (2) 1 + tan 2 tan = sec 2 (3) 2 cosec 2 = sec cosec (4) cos4 - sin4 = cos 2 (5) cot - tan = 2cot 2 (6) cot 2 A (7) cot A tan A cos 2 A cot A tan A (8) 1 cot 2 A cos ec 2 A 2cot A (9) cot 2 A 1 sec 2 A cot 2 A 1 (10) 1 sec 2cos2 sec 2 (11) 2 sec2 cos 2 sec2 (12) cos ec 2 2 cos 2 cos ec 2 31 cot 2 A 1 2cot A cos 2 cot 45 1 sin 2 2 (13) A A sin cos 1 sin A 2 2 (15) sin 8A = 8sin A cos A cos 2A cos 4A (17) cos 2 sin 2 cos3 sec cosec 2 (14) 2 (16) 1 + cos 4A = 2cos 2A(1 – 2sin2 A) tan 4 (18) cos 4x = 1 – 8 sin x cos x (20) (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0 (21) cot (22) cosec (sec - 1) – cot (1 – cos ) = tan - sin (23) tan(45 + A)– tan(45 - A) = 2tan 2A (24) tan (45 + A) + tan(45 - A) = 2 sec 2A (26) cot 3 A (28) sin 3 sin 3 cot cos3 cos3 (30) cos 5A cos 2A – cos 4A cos 3A = -sin 2A sin A (31) sin 4 cos - sin 3 cos 2 = sin cos 2 (32) 4sin A sin(60 + A)sin(60 - A) = sin 3A (33) 2 2 4cos cos cos cos3 3 3 (34) 2 2 cos cos cos 0 3 3 (35) cos2 A + cos2 (60 + A) + cos2 (60 – A) = 3/2 (36) sin2 A + sin2(120 + A) + sin2(120 - A) = 3/2 (37) (38) (39) (19) 4 tan 1 tan 2 1 6 tan tan 2 4 sin cos ec 1 cos cot 3 A 3cot A 3cot 2 A 1 sin sin sin sin sin sin cot 2 cot (25) sin 3 A cos3 A 2 sin A cos A (27) 3cos cos3 cot 3 3sin sin 3 (29) cos3 cos3 sin 3 sin 3 3 cos sin 2 (cos A – sin A)(cos 2A – sin 2A) = cos A – sin 3A sin 2 A cos 2 A 1 tan A2 2 tan 2 A 1 tan A 2 (40) sin 4 A 4 tan A 1 tan 2 A 1 tan A 2 32 2 4cos 2 A 1 2sin 2 A (41) cot 15 A tan 15 A (42) cot 15 A tan 15 A (43) tan A 30 tan A 30 (44) (2cos A + 1)(2cos A – 1)(2cos 2A – 1) = 2cos 4A + 1 (45) cos2( - ) + cos2( - ) + cos2( - ) = 1 + 2cos ( - )cos( - )cos( - ) (46) 2cos2 (45 - A) = 1 + sin 2A (47) 2cos 2x cosec 3x = cosec x – cosec 3x (49) 2tan 2x = (50) (cos x + cos y)2 + (sin x – sin y)2 = 4cos2 x y 2 (51) (cos x – cos y)2 + (sin x – sin y)2 = 4sin 2 x y 2 (52) 1 sin 2 x tan 2 x 1 sin 2 x 4 (54) (sin3x sin x)sin x (cos3x cos x)cos x 0 (55) 1 1 sin A 2sin 3 A sin 5 A sin 3 A cot 2 A. (56) tan 3 A tan A cot 3 A cot A sin 3 A 2sin 5 A sin 7 A sin 5 A (57) sin 6 (58) 3A A A A 4 cos sin sin cos ec 2 2 6 2 3 2 (59) cos 5 = 16 cos5 - 20 cos3 + 5 cos (60) cos A cos (60 – A) cos(60 + A) = ¼ cos 3A (61) cos8 A cos 5 A cos12 A cos 9 A tan 4 A sin 8 A cos 5 A cos12 A sin 9 A (62) cos3 A cos3 120 A cos3 240 A 4 cos 2 A 3 sin 2 A 1 2cos 2 A 1 2cos 2 A (48) cos A cos 2A cos 4A cos 8A = sin16 A 16sin A cos x sin x cos x sin x cos x sin x cos x sin x (53) cos3 sin 3 2cot 2 sin cos A A 3 cos 6 sin 2 A 4 cos A. 2 2 4 33 3 cos 3 A 4 (63) 2sin A cos3 A – 2sin3 A cos A = 1 sin 4A 2 (64) sin 3A cos3 A + cos 3A sin3 A = 3 sin 4A 4 (65) tan 3 cot 3 1 2sin 2 cos 2 sin cos 1 tan 2 1 cot 2 (66) 2cos cos cos ( + ) = cos2 + cos2 - sin2 ( + ). sin 2 3 cos 2 3 8cos 2 sin 2 cos 2 (67) 3 1 sin 4 A cot 2 A cos 4 A 0 4 (69) 2cos - cos 3 - cos 5 = 16cos3 sin2 (70 cos3 ( + ) sin3 ( - ) + cos3 ( + ) sin3 ( - ) + cos3 ( + ) sin3 ( - ) (68) = 3cos ( + ) cos ( + ) cos( + ) sin( - ) sin( - ) sin ( - ) 2. (71) cos2 A + cos2 B – 2cos A cos B cos(A + B) = sin2 (A + B). (72) tan 3 A cot 3 A sec A cos ec A sin 2 A 1 tan 2 A 1 cot 2 A (73) tan tan 60 tan tan 60 tan 60 tan 60 3 (74) 2sin4 - sin 10 +sin2 = 16sin cos cos 2 sin 2 3 . Prove the following numerical equalities (based on last three sections) (1) cos 80 + cos 40 – cos 20 = 0 (2) cos 130 cos 40 + sin 130 sin 40 = 0 (3) cos 5 – sin 25 = sin 35 (4) cos 70 cos 10 + sin 70 sin 10 = ½ (5) sin 65 + cos 65 = 2 cos 20 (6) sin 78 – sin 18 + cos 132 = 0 (7) cot 15 + cot 75 + cot 135 – cosec 30 = 1 (8) (9) sin 10 sin 30 sin 50 sin 70 = 1/16 1 sin 20 sin 40 sin 80 = 3 8 (10) tan 40 + cot 40 = 2sec 10 2 (11) tan 70 + tan 20 = 2 cosec 40 (12) (13) tan 6 tan 42 tan 66 tan 78 = 1 (14) cos 33 cos 57 1 (16) sin 21 cos 21 2 cos 20 2sin 2 55 1 2 sin 65 2 (15) (17) 2 34 1 cot 60 1 cos30 1 cot 60 1 cos30 cosec 10 - 3 sec10 = 4 1 cos ec10 2sin 70 1 2 3. (18) sin 70 8cos 20 cos 40 cos80 2 cos 2 10 (19) cos 2 73 cos 2 47 cos 73 cos 47 (20) tan10 tan 70 tan 50 3 (21) tan 20 + tan 40 + 3 4 3 tan 20tan 40 3 If sin A + sin B + sin C = cos A + cos B + cos C = 0, prove that sin2 A + sin2 B + sin2 C = cos2 A + cos2 B + cos2 C = 3/2 ANSWERS 1. (1). 2 5 5 , ,2 5 5 (2). 6 3 , , 2 3 3 (3). 8 2 15 8 2 15 , , 4 15 4 4 Section –VI Periodicity, Graphs of Trigonometric functions All basic trigonometric functions are periodic. The functions sin x, cos x, sec x and cosec x have periods 2n (n is any integer) with fundamental period as 2 since sin (x + 2n) = sin x cos(x + 2n) = cos x sec(x + 2n) = sec x cosec(x + 2n) = cosec x The periods of tan x and cot x are n with fundamental period as . Note that 1. The period of sin 2x is 2/2 = by using the fact that if period of & f(x) is T then period of f(x) is T/ ( 0). 2 6 . 1 3 2. Using (1) period of tan 3x is /3 while period of cot x/3 is 3. Fundamental period of sin 4. 5. 6. S1 = 6, 12, 18, 24, 30, …………………. (periods of sin x/3) and S2 = 8, 16, 24, 32, ………………… is 24 Fundamental period of sin x, cos(sin x) and sin2 x are Period of sin x + cos x is /2 Functions of the type sin x2, sin x , sin 1/x are non periodic. x x cos is 24 since least common element in the two series 3 4 Graphs Since trigonometric functions are periodic we need to trace their graphs only within periods. Let us observe the graph of sin x we observe that (1) sin x increases from 0 to 1 when x increases from 0 to /2. (2) sin x decreases from 1 to 0 when x increases from /2 to . 35 (3) sin x further decreases from 0 to -1 when x increases from to 3/2. (4) sin x finally increases from -1 to 0 when x increases from 3/2 to 2. The same pattern repeats from 2 to 4 and so on. The following graphs can easily be traced by using the graph of sin x (a) sin 3x (or sin x in general) (b) 2 sin 3x (or sin x) (c) 2sin 3x + 5 To trace y = sin 3x we will draw the graph of sin x and will divide each of the located point on x-axis by 3. Again to draw y = 2 sin 3x we will draw the graph of sin x then (a) Change (0, 1) and (0, -1) to (0, 2), (0, -2) (b) divide each point marked on x-axis by 3. Finally to draw y = 2 sin 3x + 5 we will raise the last graph by 5 units (Translation along positive y-axis) Note that graph of sin (x + ) can also be traced by graph of sin x. We just have to subtract from each point located on x-axis. The graph of cos x is easily seen to as shown in the figure. As in the case of sin x the graphs of associated functions can also be traced. Graphs of other trigonometric functions Since tan x, cot x, sec x, cosec x may approach - or their graphs have tangent lines meeting them (called asymptotes) at infinity. These functions have infinite discontinuities at n (In case of cot x and cosec x) and at (2n + 1)/2 (In case of tan x and sec x) Tan x Cot x 36 Graph of sec x Graph of cosec x can similarly be traced Solved Examples 1. Compare the graphs of y = sin 2x and y sin x 1 tan 2 x cos x 1 cot 2 x . Solution: The graph of y1 = sin 2x is continuous and smooth through out the interval [0, 2] (Indeed over the entire number line). On the contrary, the function y2 sin x 1 tan 2 x n not defined at ) 2 Also cos x 1 cot 2 x 1 tan 2 x sec x Thus, if x n then 2 and is not defined at x n (because tan x, cot x are 2 1 cot 2 x cos ecx y2 = sin x sec x + cos x cosec x Observe the following table for the function y2 (Note sin x cos x + sin x cos x = sin 2x) 37 Interval sec x cosec x y2 0, 2 sec x cosec x sin 2x , 2 -sec x cosec x 0 3 , 2 -sec x -cosec x -sin 2x 3 , 2 sec x -cosec x 0 Exercise 1. From the graph of y sin x draw the graph of following functions : (ii) (iv) y sin 3x y 3 sin 5 x (v) y 2 sin x (iii) y 3 sin 5 x (vi) (vii) y cos x (viii) y sin 2 x (i) 2. Draw the graphs of sin x + cos x [Hint: sin x + cos x = 38 y sin x 2 y 3 sin 5x 3 2 sin x ] 4