Trigonometric Functions

Trigonometric Functions TRIGONOMETRY The word 'trigonometry' is derived from the Greek words 'trigon' and ' metron' and it means 'measuring the sides and angles of a triangle'. ANGLE : Angle is a measure of rotation of a given ray about its initial point. The original ray is called the initial side and the final position of the ray after rotation is called the terminal side of the angle. The point of rotation is called the vertex. If the direction of rotation is anticlockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is negative. Quadrant : Let XOX and YOY be two lines at right angles in the plane of the paper. These lines divide the plane of the paper into four equal parts which are known as quadrants. The lines XOX and YOY are known as x-axis and y-axis respectively. These two lines taken together are known as the coordinate axes. The regions XOY, YOX, X OY and YOX are known as the first, the second, the third and the fourth quadrant respectively. SYSTEMS FOR MEASUREMENT OF ANGLES : An angle can be measured in the following systems. 1. Sexagesimal System (British System) : The principal unit in this system is degree ( º ). One right angle is divided into 90 equal parts and each part is called one degree (1º ). 23 Trigonometric Functions One degree is divided into 60 equal parts and each part is called one minute. Minute is denoted by (1'). One minute is equally divided into 60 equal parts and each part is called one second (1''). Ex.2 50' is equal to  1 (A) 1g  1 1 i. e. 360 g (B)  2    g (C)  4  (D) None of these of a complete circular turn is called a degree   Sol. 100' is equal to 1g 1 1 (°), of a degree is called a minute () and of g g 60 60 1 1 50 50' is equal to 100 = 2 Ans.[B] aminute is called a second (). FG H One right angle = 90°, 1° = 60, 1= 60 3. 30º +  2  =  2      Ans. [C] Circular System (Radian Measurement) The angle subtended by an arc of a circle whose length is equal to the radius of the circle at the centre of the circle is called a radian. In this system the unit of measurement is radian (c) As the circumference of a circle of radius 1 unit is 2 , therefore one complete revolution of the initial side subtends an angle of 2 radian. More generally, in a circle of radius r, an arc of length r will subtend an angle of 1 radian. It is well-known that equal arcs of a circle subtend equal angle at the centre. Since in a circle of radius r, an arc of length r subtends an angle whose measure is 1 radian, an arc of length  will subtend an angle whose measure Centesimal System (French System) : is Ex.1 30º 30' is equal to  41  (A)  2     61  (C)  2    º (B) 61º º (D) None of these  1 Sol. We know that, 30' =  2     1 2. IJ FG IJ K H K º  61  º º The principal unit in system is grade and is denoted by ( g ). One right angle is divided into 100 equal parts, called grades, and each grade is subdivided into 100 minutes, and each minutes into 100 seconds. i.e. (g), 1 400 1 100  radian. Thus, if in a circle of radius r, arc of r length  subtends an angle  radian at the centre, we have  =  or = r . r of a complete circular turn is called a grade of a grade is called a minute a minute is called a second and 1 100 of .  One right angle = 100g; 1g = 100 ; 1 = 100 Note : The minutes and seconds in the Sexagesimal system are different with the minutes and seconds respectively in the Centesimal System. Symbols in both systems are also different. 24 Some Important Conversion :  Radian 6 = 180º One radian = Radian = 30º 4 Radian = 60º FG 180 IJ H K º Radian = 45º Trigonometric Functions Ex.4 Find the length of an arc of a circle of radius 5 cm subtending a central angle measuring 15º. Sol. Let s be the length of the arc subtending an angle  at the centre of a circle of radius r. then ,  s r = Radian = 90º 3 2 Radian = 120º 3 5 Radian = 150º 6 5 Radian = 225º 4 2 3 Radian = 135º 4 7 Radian = 210º 6 5 Radian = 300º 3 Note : If no symbol is mentioned while showing measurement of angle, then it is considered to be measured in radians. e.g.  = 15implies 15 radian # Area of circular sector : 1 Area = r2 sq. units 2 RELATION BETWEEN RADIAN, DEGREE AND GRADE : Here, r = 5 cm, and  = 15º =  F I = GH 12 JK s = C  5 cm. 12 = FG15 H IJ 180 K C s s  = r 12 5 Ans. [C] TRIGONOMETRICAL RATIOS OR FUNCTIONS In the right angled triangle OMP , we have base (OM) = x, perpendicular (PM) = y and hypotenuse (OP) = r, then we define the following trigonometric ratios which are known as trigonometric function. sin = P H y r cos = tan = P B y x cot = sec = H B r x B H B P H cosec = P = x r x y r y Note : (1) It should be noted that sin does not mean the product of sin and . The sin is correctly read sin of angle . (2) These functions depend only on the value of the angle  and not on the position of the point P chosen on the terminal side of the angle . Where, D = angle in degree G = angle in grade R = angle in radian Fundamental Trigonometrical Identities : 1 1 (b) cos  = cos ec sec 1 cos (c) cot = tan sin (d) 1 + tan2 = sec2 or, sec2 – tan2 = 1 1 (sec – tan) = (a) sin = Ex.3 340º is equal to  (A)  9    F 17 I (B) GH 9 JK C  17   (C)  6    C  16   (D)  9    C C Sol. We know , 180º = C FG H IJ K 340º = 180 340 C = FG 17 IJ H9K (e) sin2  + cos2 = 1 (f) 1 + cot2  = cosec2 C Ans.[B] (cosec – cot) = 1 ec 25 Trigonometric Functions SIGN OF THE TRIGONOMETRIC tan   sec   1 = tan   sec   1 1  sin  (A) cos  1  sin  (C) cos  FUNCTIONS : Ex.5 Sol. (B) (D) 1  cos  sin  1  cos  sin  2 1 = 1 2 2 [ sec  – tan  = 1] 2 1 1 = 1 1 = 1 1 sin 1 sin = sec + tan = + = cos cos cos (i) If is in the first quadrant then P(x, y) lies in the first Ans.[C] Ex.6 The value of the expression 2 sin y 1 – 1  cos y + (A) 0 (C) sin y 2 sin y Sol. 1 – 1  cos y + of all the trigonometric functions are positive. 1 cos y sin y – sin y 1  cos y is equal to- (ii) If is in the quadrant then P(x, y) lies in the  quadrant. Therefore x < 0, y > 0 and hence the (B) 1 (D) cos y values sin, cosec are positive and the remaining are negative. 1 cos y sin y – sin y 1  cos y 1  cos y  sin2 y 1  cos 2 y  sin2 y = + 1  cos y sin y (1  cos y ) cos y  cos 2 y = + 0 = cos y 1  cos y quadrant. Therefore x > 0, y > 0 and hence the values (iii) If is in the quadrant then P(x, y) lies in the  quadrant. Therefore x < 0, y < 0 and hence the values of tan, cot are positive and the remaining are Ans.[D] Ex.7 If cosec  – sin  = m and sec  – cos  = n then (m2n)2/3 + (n2m)2/3 equals to (A) 0 (B) 1 (C) –1 (D) 2 Sol. cosec  – sin  = m cos 2  ...(i) sin  sin 2  ...(ii) cos  cos 2  sin 2  m×n= . = sin  cos  sin  cos  1 – sin  = sin 1 n= – cos  = cos m= negative. (iv) If is in the V quadrant then P(x, y) lies in the IV quadrant. Therefore x > 0, y < 0 and hence the values of cos, sec are positive and the remaining are negative. To be Remember : from (i) and (ii) from (i) cos2  = m . sin  or cos3  = m sin  cos  = m . (mn) = m2n Similarly sin3  = n2m since sin2  + cos2  = 1 (n2m)2/3 + (m2n)2/3 = 1 Ans.[B] 26 Trigonometric Functions But  lies in the fourth quadrant in which sin is negative. Values of trigonometric functions of certain popular angles are shown in the following table :  6 0 0 0 4 sin cos 1 tan 0 1 1  4 2 3 2 1  4  3 2 1  4 2 1 2 3 3  4 2 1 2 1 3 3  2 4 1 4 tan = 0  N.D. N.D. implies not defined The values of cosec x, sec x and cot x are the reciprocal of the values of sin x, cosx and tan x, respectively. 1 sin = 2 , sin cos cosec  =  then, 1 tan = tan = –1 1 1  2 2 2 1 cot = –1 ec = ec  –1 Ans. [A] TRIGONOMETRIC RATIOS OF ALLIED ANGLES : 12 and  13 Ex.8 The values of sin and tan if cos = If  is any angle, then  2 ±  etc. are called allied angles. lies in the third quadrant is5 5 (A)  13 and 12 (B) 5 5 and  13 12 5 12 (C) and  13 (D) None of these 13 Sol. We have cos2 + sin2= 1  sin = 1 cos 2 In the third quadrant sin is negative, therefore sin = 1 cos 2 1  sin = FG H sin cos then, tan = 12 13 IJ K 2 * Trigonometric Ratios of ( –  ) : Let be an angle in the standard position in the  quadrant. Let its terminal side cuts the circle with centre ‘O’ and radius r in P (x, y). Let P(x, y) be the point of intersection of the terminal side of the angle – in the standard position with the circle. Now MOP = MOP (numerically) and P & P have the same projection M in the x - axis 5 13  tan =  3 ± ,  ± , ± , 2 2  OMP OMPx = xand y = – y 5 13 13 5 12 12 Ans.[A] 3 Ex.9 If sec = 2 , and <  < 2. Then the value of 2 1 1 ec ec is1 (A) – 1 (B)  2 (C) – 2 (D) 1 Sol. If sec = sin = 2 1 cos 2 or, = cos = 1 1 = 2 1 2 , 1 2 –y  sin (– ) = y = = – sin . r r x x cos (– ) = = = cos  r r 27 Trigonometric Functions tan (– ) = cot (– ) = y y = – = – tan  x x *  2   sin  –   = cos  , r r sec (– = = = sec. x x    cosec  –   = sec  2    cot  –   = tan , 2   r cosec (– ) = y = – y = – cosec  Similarly if is in the other quadrants then the above results can also be proved. * * Trigonometric Ratios of –     cos  –   = sin , 2    sec  –   = cosec  2    2   Trigonometric Ratios of         quadrant. Let its terminal side cuts the circle with sin  2    = cos , centre ‘O’ and radius r at P (x, y). Let P(x, y) be cosec  2    = sec        cot  2    = – tan ,   the point of intersection of the terminal side of the angle – with the circle. Let M and Mbe the *    tan  2    = – cot ,     cos  2    = – sin ,     sec  2    = – cosec    Trigonometric Ratios of ( + ) Similarly we can easily prove the following results. Since OMPOMP, x= – x, y= y y y  sin (– ) = = = sin . r r x x cos (– ) = = – = – cos. r r  2    tan  –   = cot  Similarly we can easily prove the following results. Let be an angle in the standard position in the projections of P and Prespectively in the x-axis.  Similarly we can easily prove the following results. x x = = – cot . y –y r  2   Trigonometric Ratios of  –   : sin (+ ) = – sin , tan (+ ) = tan  cosec (+ ) = – cosec  cos (+ ) = – cos , cot (+ ) = cot , sec (+ ) = – sec  *  3  –  2   Trigonometric Ratios of  Similarly we can easily prove the following results.  3  –   = – cos , 2    3  cosec  –   = – sec   2  sin   3  –  2   cot  y y tan (– ) = = = – tan . x –x x x cot (– ) = = – = – cot . y y r r sec (– ) = = = – sec . x –x r r cosec (– ) = y = y = cosec . * = tan ,  3  –   = cot , 2    3  cos     = – sin ,  2  tan   3  –   = – cosec , 2   sec   3    2   Trigonometric Ratios of  Similarly we can easily prove the following results.  3     = – cos , 2    3     = – cot ,   2  sin  tan  3     2  sec  = cosec , cos  3    = sin ,  2   3  cot     = – tan ,  2   3  cosec     = – sec   2  28 Trigonometric Functions Think, and fill up the blank blocks in following table. 0 sin 0 cos 1 tan 0  6 1 2  4 1 2  3 3 2 3 2 1 3 1 2 1 2 1  2 2 3 5 6 (C) (B) – 1 2 1 2 3 2 5 3 11 6 2 N .D. Sol. 2 4 3 0 3 1 7 6 1 Ex.10 sin 315º = (A)  cos ec(2  ).cos(2  ) tan(  / 2  ) sec(  / 2  ).cos .cot(   ) cos ec  .cos  (  cot ) = (  cos ec ).cos .cot  = 1 Ans.[D] (D) None of these TRIGONOMETRIC FUNCTIONS: Sol. sin 315º  sin (270º + 45º)  – cos 45º [  sin (270º +  ) = – cos]  – 1 Ans.[B] 2 (i) y = sin x Domain : x  R Range : y  [–1, 1] Ex.11 cos 510º cos 330º + sin 390º cos 120º = (A) 2 (B) – 1 (C) 0 (D) 1 2 (ii) y = cos x Sol. cos 510º cos 330º + sin 390º cos 120º Domain : x  R Range : y  [ – 1, 1] = cos (360º + 150º) cos (360º – 30º) + sin(360º + 30º) cos(90º + 30º) = cos 150º cos 30º + sin 30º(–sin 30º) = cos (180º – 30º) = – cos 30º F 3I GH 2 JK F 3I GH 2 JK – – 1 4 1 4 (iii) y = tan x    2 Domai : xR – (2n  1)  , n Range : y  R 3 1 – =–1 Ans.[B] 4 4 cos ec(2  ).cos(2  ) tan(  / 2  )  Ex.12 sec(  / 2  ).cos .cot(   ) =– (A) 2 (C) 4 0 (B) – 1 (D) 1 29 Trigonometric Functions (iv) y = cot x Domain : x  R – {n}, n  Range : y  R cot A cot B  1 (f) cot (A ± B) = cot B  cot A (g) sin (A + B + C) = sin A cos B cos C + sin B cos A cos C + sin C cos A cos B – sin A sin B sin C (h) cos (A + B + C) = cos A cos B cos C – cos A sin B sin C – sin A cos B sin C – sin A sin B cos C (v) y = cosec x Domain : x  R – {n}, n  Range : y  (,  1]  [1, ) (i) tan (A + B + C) tan A  tan B  tanCtan A tan B tan C = 1  tan A tan B  tan B tan C tan C tan A . (j) tan (1 + 2 + 3 + ... + n) = S1  S3  S 5  ...... 1  S2  S 4  ....... where Si denotes sum of product of tangent of angles taken i at a time (vi) y = sec x   2 Domain : xR– (2n  1)  , n  Range : y  (,  1]  [1, ) Ex.13 Prove that (i) sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = cos (A – B)    3  (ii) tan  4    tan  4    = –1     Sol. (i) Clearly sin (45º + A) cos (45º – B) + cos (45º + A) sin (45º – B) = sin (45º + A + 45º – B) = sin (90º + A – B) = cos (A – B) 3      (ii) tan  4    × tan  4        TRIGONOMETRIC FUNCTIONS OF = 1  tan  1  tan  × 1  tan  1  tan  =–1 SUM OR DIFFERENCE OF TWO ANGLES : FORMULAE FOR PRODUCT INTO (a) sin (A ± B) = sinA cosB ± cosA sinB SUM OR DIFFERENCE (b) cos (A ± B) = cosA cosB  sinA sinB CONVERSION (c) sin²A  sin²B = cos²B  cos²A = sin (A+B). sin (A B) We know that, (d) cos²A  sin²B = cos²B  sin²A = cos (A+B). cos (A B) sin A cos B – cos A sin B = sin (A – B) ......(ii) tan A  tan B (e) tan (A ± B) = 1  tan A tan B sin A cos B + cos A sin B = sin (A + B) ......(i) cos A cos B – sin A sin B = cos (A + B) .....(iii) cos A cos B + sin A sin B = cos (A – B) .....(iv) 30 Trigonometric Functions Adding (i) and (ii), 2 sin A cos B = sin (A + B) + sin (A – B) Subtracting (ii) from (i), 2 cos A sin B = sin (A + B) – sin (A – B) Adding (iii) and (iv), 2 cosA cos B = cos (A + B) + cos (A – B) Subtraction (iii) from (iv). 2 sin A sin B = cos (A – B) – cos (A + B) Ex.14 Prove that sin 5A + sin 3A = 2sin 4A cos A Sol. L.H.S. sin 5A + sin 3A = 2sin 4A cos A = R.H.S. CD C D cos ] 2 2 [ sin C + sin D = 2 sin Ex.15 Find the value of 2 sin 3 cos  – sin 4 – sin 2 Sol. 2 sin 3 cos  – sin 4 – sin 2 = 2 sin 3 cos  – [2 sin 3 cos  ] = 0 Ex.16 Prove that sin 8 cos   sin 6 cos 3 Formulae : (i) cos 2 cos   sin 3 sin 4 = tan 2 (a) 2 sin A cos B = sin (A + B) + sin (A – B) (ii) tan 5  tan 3 = 4 cos 2 cos 4 (b)2 cos A sin B = sin (A + B) – sin (A – B) tan 5  tan 3 Sol. (i) (c) 2 cos A cos B = cos (A + B) + cos (A – B) (d)2 sin A sin B = cos (A – B) – cos (A + B) 2 sin 8 cos   2 sin 6 cos 3 2 cos 2 cos   2 sin 3 sin 4 = 2 sin 2 cos 5 = 2 cos 5 cos 2 = tan 2 FORMULAE FOR SUM OR (ii) DIFFE RENCE INTO PRODUCT = CONVERSION Ex.17 (cos  + cos  )2 + (sin  + sin )2 = We know that, sin (A + B) + sin(A – B) = 2 sin A cos B ......(i) Let A + B = C and A – B = D then A = 2 and B = (a) sin C + sin D = 2 sin FG H 2 IJ .cos FG K H 2 IJ K IJ .sin FG IJ H 2 K 2 K F I F I (c) cos C + cos D = 2 cos GH 2 JK .cos GH 2 JK F IJ .sin FG IJ (d)cos C – cos D = 2 sin GH H 2 K 2 K FG H Formulae for sum or difference into product conversion FG    IJ H 2 K F I (D) 4cos2 GH 2 JK (B) 4cos2 2 similarly other formula are, (b)sin C – sin D = 2 cos FG    IJ H 2 K F I (C) 4sin2 GH 2 JK (A) 4sin2 Sol. (cos  + cos )2 + (sin  + sin )2 2 Substituting in (i), * tan 5  tan 3 sin 5 cos 3  sin 3 cos 5 = tan 5  tan 3 sin 5 cos 3  sin 3 cos 5 sin 8 = 4 cos2 cos 4 sin 2        = 2 cos 2  . cos 2  +               . cos  2 sin   2   2   FG H F + 4sin2 GH = 4cos2 2 2 IJ .cos2 FG K H IJ .cos2 FG K H –  2 2 2 IJ K IJ K     2        sin   2   2  2 = 4 cos2  2  . cos       = 4 cos 2  2    Ans.[D] 31 Trigonometric Functions MULTIPLE AND SUB-MULTIPLE ANGLES : (i) sin 2 = 2 sin  cos  = (ii) cos 2 = cos2  = 1 – 2 sin2  = (iii) tan 2 = – sin2   AB B 2 sin cos   A  2 A B  2  = tan = tan cot 2  A  B B  2 2  2 sin 2 sin 2      2 tan 1 tan2 = 2 cos2  – 1 Sol. 3 tan   tan 3  (viii) cos /2 = (ix) tan /2 = 1  3 tan 2  1 cos 2 1 cos 2 1 cos 1 cos 1 cos = sin 1 = 1 FG IJ H 2K FG IJ H 2K FG IJ LM H 2K N FG IJ LM H 2K N 2 = sin = 1 cos 2 = 2 Ex.18 Prove that sin 2A (i) 1  cos 2A = tan A (ii) tan A + cot A = 2 cosec 2 A 1  tan 2 A (ii) L.H.S. tan A + cot A = =2 tan A  1  tan 2 A  2    2 tan A  = sin 2 A = 2 cosec 2 A   1  cos A  cos B  cos( A  B) (iii) L.H.S. 1  cos A  cos B  cos( A  B) FG IJ H 2K FG IJ H 2K FG IJ H 2K FG IJ H 2K OP Q OP = tan FGH 2 IJK Q 3     Ans.[D] Ex.20 The value of 1  cos 8  1  cos 8  5    1  cos   8 1  cos A  cos B  cos( A  B) A B (iii) 1  cos A  cos B  cos( A  B) = tan cot 2 2 2 sin A cos A sin 2A Sol. (i) L.H.S. = = tan A 1  cos 2A 2 cos 2 A 7    1  cos   8 is - 1 2 1 (C) 8 (A) Sol.  8 1 2 (B) cos (D) 2 2 3    3        1  cos     1  cos  1  cos       8   8  3      3      8   = 1  cos 8  1  cos 8  1  cos 8  1  cos 8    2  = 1  cos 8  2 sin 2  A A  sin  sin  B    A 2 2   = tan 2  A A    cos 2  cos 2  B      (B) sin 2 1 1 2 A A A   2 sin sin  B  2 2 2  = A A A   2 cos 2  2 cos cos  B  2 2 2  FG IJ H K F I (D) tan GH 2 JK (A) cot 2 (v) cos 3 = 4 cos3  – 3 cos  (vii) sin /2 = = FG IJ H K F I (C) cos GH 2 JK 1 tan2 2 tan 1 tan2 (iv) sin 3 = 3 sin  – 4 sin3  (vi) tan 3 = 1 1 Ex.19 1 tan2  2 3   1  cos   8  = 1 4   3    2  1  cos   2  1  cos   4  4  = 1 4    1  cos   4 = 1 4 3    1  cos   4   1  1  1 1  1 1 1   1   =   =   2   2 2 4 8   Ans.[C] 32 Trigonometric Functions Ex.21 The value of sin 20° sin 40° sin 60° sin 80° is3 (A) 8 3 (C) 16 1 (B) 8 IDENTITIES We have certain trigonometric identities like, (D) None of these sin2 + cos2 = 1and 1 + tan2 = sec2 etc. Sol. sin 20° sin 40° sin 60° sin 80° = = = = = = 3 2 3 2 3 2 3 8 3 8 3 8 sin 20° sin (60° – 20°) sin (60° + 20°) sin 20° (sin2 60° – sin2 20°) sin 20° ( given condition among them and they are called conditional identities. 3 – sin2 20°) 4 If A, B, C denote the angle of a triangle ABC, (3 sin 20° – 4 sin3 20°) then the relation A + B + C =  enables us to establish many important identities involving sin 60° trigonometric ratios of these angles. 3 3 = 16 2 . (I) If A + B + C = , then A + B =  – C, B + C Ans.[C] =  – A and C + A =  – B 1 (II) If A + B + C = , then sin(A + B) = sin( – C) sin  .sin(60º –  ).sin (60º +  ) = sin 3  4 = sinC 60º [sin 20º sin (60º – 20º) similarly, sin (B + C) = sin(  – A) = sin A and sin (60º + 20º)] 1 sin (C + A) = sin ( – B) = sin B 1  3  2 (III) If A + B + C = , then cos(A + B) 3   = = sin 60º  4 sin 60 º  =   4  2  16 Ex.22  cos4 8 + 3 cos 4 8 equals to (A) 1/2 (B) 1/4 Sol. =  cos4 3 +cos4 + 5 cos4 + 8 7 cos4 8 2 = + 5 cos 4 (D) 3/4 + 7 cos4 2 1   3    2 cos 2    2 cos 2   2  8 8      2 2 1   3    1  cos    1  cos   2  4 4      2 2 1  1   1   1 1    1 –  21     = = 2  2 2     2  Ans.[C] = cos( – C) = – cos C similarly, cos (B + C) = cos( – A) = – cosA and cos (C + A ) = cos ( – B) = – cos B (C) 3/2 8 8 8 8  3  3   = cos4 + cos4 + cos4 + cos4 8 8 8 8  4  4 3  = 2  cos 8  cos 8  = Such identities are identities in the sense that they hold for all value of the angles which satisfy the Alternate : By direct formula  sin CONDITIONAL TRIGONOMETRICAL (IV) If A + B + C = , then tan(A + B) = tan( – C) = – tan C similarly, tan (B + C) = tan(  – A) = – tanA and, tan (C + A ) = tan (  – B) = – tan B (V) If A + B + C = , then 2 = 3 2 FG H FG H FG H 2 2 2 2 IJ K IJ K IJ K 2 and FG H2 FG H2 FG H2 2 IJ 2K IJ K IJ 2K 2 = 2 2 C and 2 FG IJ H 2K FG IJ H 2K FG IJ H 2K 33 All problems on conditional identities are broadly divided into the following four types: (I) Identities involving sines and cosines of the multiple or sub-multiples of the angles involved. (II) Identities involving squares of sines and cosines of the multiple or sub-multiples of the angles involved. (III) Identities involving tangents and cotangents of the multiples or sub-multiples of the angles involved. (IV) Identities involving cubes and higher powers of sines and cosines and some mixed identities. 1. TYPE I : Identities involving sines and cosines of the multiple or sub-multiple of the angles involved. Working Method : 3. Trigonometric Functions Type III : Identities for tan and cot of the angles Working Method : Step – 1 Express the sum of the two angles in terms of third angle by using the given relation (A + B + C = ). Step – 2 Taking tangent or cotangent of the angles of both the sides. Step – 3 Use sum and difference formulae in the left hand side. Step – 4 Use cross multiplication in the expression obtained in the step 3 Step – 5 Arrange the terms as per the result required. Conditional trigonometrical identities Step – 1 Use C & D formulae. Step – 2 Use the given relation (A + B + C = ) in the expression obtained in step -1 such that a factor can be taken common after using multiple angles formulae in the remaining term. Step – 3 Take the common factor outside. Step – 4 Again use the given relation (A+ B + C = ) within the bracket in such a manner so that we can apply C & D formulae. Step – 5 Find the result according to the given options. 2. TYPE II : Identities involving squares of sines and cosines of multiple or sub-multiples of the angles involved. Ex.23 If A + B + C = , then sin 2A + sin 2B+ sin 2C = (A) 4sin A sin B cos C. (B) 4sin A sin B sin C. (C) 4cos A sin B sin C. (D) None of these Sol. sin 2A + sin 2B + sin 2C = 2 FG H 2 IJ FG K H 2 IJ K 2C = 2sin(A + B).cos(A – B) + sin2C Working Method : = 2 sin ( – C).cos(A – B) + sin 2C [  A + B + C = , A + B =  – C Step – 1 Arrange the terms of the identity such that either sin2A – sin2B = sin(A + B). sin(A – B)  or cos2A – sin2B = cos(A + B).cos(A – B) can be used. = 2 sinC [cos(A – B) + cosC] Step – 2 Take the common factor outside. Step – 3 Use the given relation (A + B + C = ) within the bracket in such a manner so that we can apply C & D formulae. Step – 4 Find the result according to the given options. sin(A + B) = sin ( – C) = sinC] = 2sinC cos (A – B) + 2 sinC cosC = 2 sinC [cos(A – B) – cos(A + B)] [  cos(A – B) – cos (A + B)= 2 sinA.sinB, By C & D formula] = 2 sinC [2sinA sinB] = 4 sinA sinB sinC Ans.[B] 34 Trigonometric Functions Ex.24 If A + B + C = , then tanA + tanB + tanC = (A) cotA.tanB.tanC (B) tanA.cotB.tanC (C) tanA.tanB.tanC (D) None of these Sol. A + B + C =      B  C A A A cos = 2 sin cos  2  2 2 2 A BC cos = cos 2 2 2 sin A+B=–C or A = B – C tan(A + B) = tan ( – C) But A + B + C =  tan C 1 tan A tan B Therefore 2B =   B = /2 Ans.[A] tanA + tanB = – tanC + tanA.tanB.tanC tanA + tanB + tanC = tanA.tanB.tanC Ans.[C] Ex.25 If A + B + C = 3 , then 2 THE GREATEST AND LEAST VALUE OF THE EXPRESSION [a sin + b cos] cos 2A + cos 2B + cos 2C = Let a = r cos ...(1) (A) 1 – 4 cos A cos B cos C (B) 4 sin A sin B sin C and b = r sin  ...(2) (C) 1 + 2 cos A cos B cos C (D) 1 – 4 sin A sin B sin C then a2 + b2 = r2 Squaring and adding (1) and (2)  Sol. cos 2A + cos 2B + cos 2C = 2 cos  A+B+C= 2 a sin + b cos = r (sin cos + cos sin) = r sin(  + ) = 2cos (A + B) cos (A – B) + cos 2C  3   C   2  2 or, But –1  sin   1 cos (A – B) + cos 2C so –1  sin( + )  1 then –r  r sin( + )  r 3 2 hence, = –2 sin C cos (A – B) + 1 – 2 sin 2 C then the greatest and least values of a sin + b cos are a2 = 1 – 2sin C [cos (A – B) + sin C)  3  = 1 – 2sin C cos ( A  B)  sin 2  A  B      respectively and a2 b2 The greatest and least value of the expression [a sin + b cos] = 1 – 2 sin C [cos (A – B) – cos (A + B)] = 1 – 4 sin A sin B sin C b2  a sin + b cos  Ans.[D] Ex.26 In any triangle ABC, sin A – cos B = cos C, then angle B is (A) 2 (B) 3 (C) (D) 4 Ex.27 The maximum value of 3 sin + 4 cos is6 (A) 2 (C) 4 Sol. We have, sin A – cos B = cos C sin A = cos B + cos C A A 2 sin cos = 2cos 2 2 2 sin  A A cos = 2 cos 2 2 A+B+C= Sol. cos 2 A 2 2 (B) 3 (D) 5 25  3 sin + 4 cos  25 [By the standard results] or, – 5  3 sin + 4 cos  5 cos 2 so the maximum value is 5. Ans.[D] 35 Trigonometric Functions MISCELLANEOUS POINTS (A) Some useful Identities : (a) tan (A + B + C) = = 1 tan A. tanB (b) tan = cot  – 2 cot 2 = (c) tan 3 = tan.tan(60º – ).tan (60º + ) (d) tan(A + B) – tanA – tanB = tanA.tanB.tan(A + B) 1 sin 3 4 1 (f) cos  cos (60º – ) cos (60º + ) = cos 3 4 (e) sin  sin (60º – ) sin (60º + ) = sin (a) sin  + sin ( + ) + sin( + 2) + .......... + to n terms ;   2n Series FG IJ + cos FG 3 IJ H K H 14 K F I 1 (A) tan GH 14 JK 2 F I 1 (C) cot GH 14 JK 2 Sol. Here cos S 14 , 2 14 FG 5 IJ H K + cos 14 (B) FG IJ H K (D) None of these and n = 3. LM   FG 3  1IJ FG 2 IJ OP sinFG 2  3 IJ N14 H 2 K H 14 K Q H 14 2 K F 2  1IJ sinG H 14 2 K Ans.[C] 1 2 2 2 3 1 2 2 1 2 2 2 4 2 5 4 4 –1 Trig. Function Domain Range sin  R [–1, 1] cos  R [–1, 1] tan  R – {2n + 1) /2, n  z} (, ) or R cosec  R – {n, n  z} (–, –1]  [1, ) sec  cot  R – ({2n + 1) /2, n  z} (–, –1]  [1, ) R – {n, n  z} (–, ) = R (E) An Increasing Product series : p = cos. cos 2 . cos 22 ..... cos (2n–1 )=  sin2n  , if   n  n  2 sin  1, if   2k  1, if   (2k  1)   = 1 cos 14 2 FG IJ H K 1  cot 2 14 (D) Domain and Range of Trigonometrical Function +..... + to n terms   n  1     n   cos     sin    2     2   = ;   2n  sin   2 S 4 tan (b) cos  + cos (  +  ) + cos ( + 2 ) Ex.28 cos 14 2 2 cos LM FG n 1IJ OP LM FG n IJ OP N H 2 K QN H 2 K Q F I sinG J H 2K FG 6 IJ H 14 K F I 2 sinG J H 14 K sin (C) Sine, cosine and tangent of some angle less than 90º. 15º 18º 22½º 36º (B) Some useful series : = FG 3 IJ sinFG 3 IJ H 14 K H 14 K FI 2 sinG J H 14 K 1 F IJ sinG H K = F I sinG J H 14 K 2 cos (F) Continued sum of sine & cosine series : (i) sin   sin    +sin   2  +..... sin   n  1   sin 36 (ii) cos   cos     cos   2  ..... Trigonometric Functions Sol. tan 9° + tan 81° – (tan 27° + tan 63°) (tan 9° + cot 9°) – (tan 27° + cot 27°)     sin 9º cos 9º sin 27 º cos 27 º =  cos 9 º  sin 9º  –  cos 27º  sin 27º       cos   n  1   1 1 = sin 9º cos 9º – cos 27 º sin 27º (F) Conversion 1 radian = 180º/ = 57º 17 45 and 1º = 180 = 0.01475 radians (approximately) (G) Basic right angled triangle are (pythogerian Triplets) 3, 4, 5 ; 5, 12, 13; 7, 24, 25; 8, 15, 17;9,40, 41;11, 60, 61; 12, 35, 37; 20, 21, 29 etc. (H) Each interior angle of a regular polygon of n sides = 2 2 2 2 = sin 18 – sin 54  = sin 18  – sin 36 º sin n 2 180 degrees n = = 24 5 1 –  5  1 5  1  2 4 = 8  ( 5  1)( 5  1)  5 1   16 =4 4 Ans.[D] n Ex.30 cos3x. sin2x =  am sin mx is an identity in x. m 1 Then 3 (A) a3 = 8 , a2 = 0 (C)  am = (B) n = 5, a1 = 3 4 1 4 (D) All the above Sol. cos3x. sin 2x = cos 3 x  3 cos x .sin 2x 4 = Ex.29 tan 9° – tan27° – tan 63° + tan 81° is equals to (A) 0 (B) 1 (C) –1 (D) 4 1 8 3 (sin 5x – sin x) + 8 (sin 3x + sin x) 3 1 1 sin x + sin 3x + sin 5x. 8 8 4 3 1  n = 5, a1 = , a2 = 0 , a3 = , 8 4 1 a4 = 0 , a5 = Ans.[D] 8 = 37 Trigonometric Functions TRIGONOMETRIC EQUATIONS GENERAL SOLUTION OF SOME STANDARD TRIGONOMETRIC An equation involving one or more trigonometric ratios of an unknown angle is called a trigonometric EQUATIONS : (i) If sin  = sin   = n  + (1)n  equation. TRIGONOMETRIC EQUATION : where     ,   , n  .  2 2 SOLUTION OF TRIGONOMETRIC (ii) If cos  = cos    = 2n ±  where   [0, ], n  . A solution of trigonometric equation is the value of (iii) If tan  = tan    = n +  the unknown angle that satisfies the equation.   where    ,  , n  .  3 9 11 1  2 2 e.g. if sin =  = , , , , ........... 4 4 4 4 (iv) If sin²  = sin²   = n  ± , n  . 2 Thus, the trigonometric equation may have infinite (v) If cos²  = cos²    = n  ± , n  . number of solutions (because of their periodic nature) (vi) If tan²  = tan²    = n  ± , n  . and can be classified as : [Note:  is called the principal angle] EQUATION : (i) Principal solution (ii) General solution. SOME IMPORTANT DEDUCTIONS : (i) sin = 0   = n, n  Principal solutions : The solutions of a trigonometric equation which lie (ii) sin = 1 in the interval [0, 2) are called Principal solutions. 1 2 e.g. Find the Principal solutions of the equation sinx = . Solution :  sinx = 1 2  there exists two values 5 6 which lie in [0, 2) and whose sine is 1 2  Principal solutions of the equation sinx = 1 2 i.e.  6  and (iii) sin = – 1  (iv) cos = 0  (v) cos = 1 (vi) cos = – 1 (vii) tan = 0    Ex.31 Solve sin  = 5 are 6 , 6 General Solution :  Sol.  sin  =  , n 2   = (4n – 1) , n  2   = (2n + 1) , n 2  = (4n + 1)  = 2n, n  = (2n + 1), n  = n, n  3 . 2 3 2   = n + (– 1)n  sin = sin  3  3 , n  The expression involving an integer 'n' which gives 1 all solutions of a trigonometric equation is called Ex.32 The general solution of cos = is – 2 General solution.   (A) 2n ± ; n  I (B) n ± ; n  I 6 6 General solution of some standard trigonometric   equations are given below. (C) 2n ± ; n  I (D) n ± ; n  I 3 3 38 1 2 Sol. If cos  = ,  = 2n ± 3 FG IJ H K or cos = cos 3 ;nI Ex.33 Solve : sec 2 = – Sol.  sec 2 = – Ans.[C] 2 3 2 Trigonometric Functions Sol. If (sin 5 + sin ) + sin3 = 0 or, 2 sin 3 cos 2 + sin 3 = 0 or, sin 3 (2 cos2 + 1) = 0 Case I sin 3 = 0  3 = n ; n  I n 3 Case II  3 5 3  cos2 = cos 6 2 5 5  2 = 2n ± , n   = n ± , n  12 6 = ;nI  cos2 = – Ex.34 Solve tan = 2 Sol.  tan = 2 ............(i) Let 2 = tan  tan = tan   = n + , where  = tan–1(2), n Ex.35 Solve cos2 = 1 Sol.  cos  = 2 2  cos2 = cos2 1 2 2cos2 + 1 = 0   cos2 = cos    4 Ex.36 Solve 4 tan2 = 3sec2 Sol.  4 tan2 = 3sec2 .............(i) 2  3  = m ± 1 2 3 ;mI So the general solution of the given equation is  = n 3 and  = m ± 3 where m, n  I Ans.[B] Ex.38 If 2cos2  + 3sin = 0, then general value of  is – 2  1  cos  =    2   = n ± , n  4 2 cos2 = (A) n + (–1)n  ;nI 6  ;nI 6  (C) n + (–1)n+1 ; n  I 6 (B) 2n ± (D) None of these 2  For equation (i) to be defined  (2n + 1) , n  Sol. If 2cos  + 3sin =0 2  2(1 – sin2) + 3sin = 0  equation (i) can be written as:  2sin2 – 3sin – 2 = 0 2 3 4 sin    2sin2 – 4sin + sin – 2 = 0 =  (2n + 1) , n  2 2 cos  2 cos   2sin(sin – 2) + (sin – 2) = 0  cos2  0  (sin – 2) (2sin + 1) = 0 2  3 Case I  4 sin2 = 3  sin2 =  2    If sin  – 2 = 0 sin  = 2   Which is not possible because –1  sin   1  sin2 = sin2   = n ± , n  3 3 Case II Ex.37 If sin  + sin 3 + sin 5 = 0, then the general If 2sin + 1 = 0 value of  is – n m , ; m, n  I 6 12 n  (B) , m ± ; m, n  I 3 3 n  (C) , m ± ; m, n  I 3 6 (A) (D) None of these  sin  = – 1 2 FG IJ H K F I   = n + (–1)n+1 GH JK 6   FG IJ H K or, sin = sin 6 = n + (–1)n 6 ;nI ;nI Ans.[C] 39 Trigonometric Functions Ex.39 If cos 3x = –1, where 0º  x  360º, then x = (A) 60º, 180º, 300º (B) 180º (C) 60º, 180º (D) 180º, 300º Sol. If cos 3x = – 1 = cos (2n + 1) or, 3x = (2n + 1)  i.e., x = 3 , , x = (2n + 1) 5 3 3 Ans.[A] Ex.40 If sin 3 = sin , then the general value of  is (A) 2n, (2n + 1) (C) n, (2n + 1) 3 3 (B) n, (2n + 1) then  = Ex.42 Solve 2 cos2x + 4cosx = 3sin2x Sol.  2cos2x + 4cosx – 3sin2x = 0  2cos2x + 4cosx – 3(1– cos2x) = 0  5cos2x + 4cosx – 3 = 0    2  19    5     cos x    4    2  19   cosx    =0...(ii) 5     cosx  [– 1, 1]  x  R (D) None of these Sol. sin 3 = sin or, 3 = m + (–1)m For (m) even i.e. m = 2n, 2n 2 Type - 2 Trigonometric equations which can be solved by reducing them in quadratic equations.  cosx   2  19 5  2  19 5  equation (ii) will be true if cosx = = n and for (m) odd i.e. m = (2n + 1) or,  = (2n +1) Ans.[B] 4 TYPES OF TRIGONOMETRIC EQUATIONS : Type-1 Trigonometric equations which can be solved by use of factorization.  cosx = cos, where cos =  2  19 5   2  19   , n   x = 2n ±  where  = cos–1    5  Type-3 Trigonometric equations which can be solved by transforming a sum or difference of trigonometric ratios into their product. Ex.41 Solve (2sinx – cosx) (1 + cosx) = sin2x. Ex43. Solve cos3x + sin2x – sin4x = 0 Sol.  (2sinx – cosx) (1 + cosx) = sin2x Sol. cos3x + sin2x – sin4x = 0  (2sinx – cosx) (1 + cosx) – sin2x = 0  cos3x + 2cos3x.sin(– x) = 0  (2sinx – cosx) (1 + cosx) – (1 – cosx) (1 + cosx) = 0  cos3x – 2cos3x.sinx = 0  (1 + cosx) (2sinx – 1) = 0  cos3x (1 – 2sinx) = 0  1 + cosx = 0 or 2sinx – 1 = 0  cos3x = 0 or 1 – 2sinx = 0 1  cosx = – 1 or sinx = 2  1  3x = (2n + 1) , n  or sinx =  2 2  x = (2n + 1), n or sin x = sin 6  x = n + (– 1) n  ,n 6  x = (2n + 1)  n + (–1)n or x = n + (–1)n  , n  6  solution of given equation is  Solution of given equation is (2n + 1), n  or  , n  6  6 , n  (2n + 1)  , n  6 or n + (–1)n  , 6 n 40 Type-4 Trigonometric equations which can be solved by transforming a product of trigonometric ratios into their sum or difference. Trigonometric Functions Step III Using the substitution in step - II, the equation reduces r cos( – ) = c  c  cos(  – ) = cos (say) r cos( – ) = Step IV Solve the equation obtained in step III by using the formula. General solution of Trigonometrical equation acos + bsin = c Ex.44 Solve sin5x.cos3x = sin6x.cos2x Sol.  sin5x.cos3x = sin6x.cos2x  2sin5x.cos3x= 2sin6x.cos2x  sin8x + sin2x = sin8x + sin4x Ex.45 General solution of equation  sin4x – sin2x = 0  2sin2x.cos2x – sin2x = 0  sin2x (2cos2x – 1) = 0  sin2x = 0 or 2cos2x – 1 = 0  2x = n, n  or cos2x = n , n  or 2   x = n ± , n 6 x=  3 , n  4 + 4 (C) 2n ± 4 + – is – ;nI 6 6 6 ;nI ;nI (D) None of these Sol. 3 cos  + sin  = 2 ....(1) this is the form of a cos + bsin = c  Solution of given equation is n , n  or 2 (A) n ± (B) 2n ± 1 2 2x = 2n ± 3 cos  + sin  = 2  where a = n ± 6 , n 3 ,b = 1 and c = Let a = r cos, and b = rsin Type 5. i.e., GENERAL SOLUTION OF TRIGONOMETRICAL EQUATION : then r = 2 and tan = acos  + bsin  = c Consider a trigonometrical equation acos  + bsin = c, where a, b, c  R and | c |  To solve this type of equation, first we reduce them in the form cos = cos or sin = sin. Algorithm to solve equation of the form acos + bsin = c Step I Step II Obtain the equation acos + bsin = c Put a = r cos and b = rsin, where r =  FG b IJ H K = tan–1 a and tan = 2 b i.e. a = rcos  and 1 = rsin 3 1 3 , so  = 6 Substituting a = 3 = rcos and b = 1 = r sin  in the equation (1) so,r [cos cos  + sin sin] = or, r cos( – ) = or, 2cos FG H F or,cos GH FG H or, cos or,  –  6 = 2n ± 2 2 IJ = 2 6K IJ = 1 6K 2 IJ = cos FG IJ H 4K 6K = 2n ± 4 + 6 4 ;nI ;nI Ans.[B] 41 Trigonometric Functions Ex.46 The number of solutions of the equation 5 sec – 13 = 12 tan in [ 0 , 2 ] is (A) 2 (B) 1 (C) 4 (D) 0 Sol. 5 sec – 13 = 12 tan or, 13 cos  + 12 sin  = 5 13 or, 5 =  As cos–1 5 313 , 13 5 313 5 313 + 313 > cos–1 + 13 313 313 Sol. sin    (D) None of these    Sol. We have, tan  2 sin  = cot  2 cos        sin  = r  + – cos , r  Z 2 2 2 sin  + cos  = (2r + 1), r  Z 2 sin  +    tan  2 sin  = tan  2  2 cos  1     cos    4  = 1 2 cos  = 2r  1 2 2 4 < 2n n +  4 < 2r  1 ,rZ 2 FG H n IJ K 3 for n > 1 4 1 Similarly for n = 8, sin  (C)  = 2r + and  = 2r, r  Z 2  1 sin IJ = n K 22 n < 2 (D) 4  n  8 or, 2< n  2 2 If n = 1, L.H.S. = 1, R.H.S. = 1/2  ,rZ 2 (B)  = 2r, r  Z  since FG H 2 2 or, 4 < n  8. (A)  = 2r +  or, sin or, tan  2 sin  = cot  2 cos  is -    + cos = 2n 2n , Ex.47 The general solution of  1 as it does not satisfy the given equation. Therefore  = 2r, r  Z Ans.[B] Ex.48 Let n be positive integer such that (C) 6  n  8 13 cos–1 then  [ 0, 2], when n = 0 (0ne value, taking positive sign) and when n = 1 (One value, taking negative sign.) Ans.[A]  or – 2 2     – = 2r ± , r  Z 4 4     = 2r + + ,rZ 4 4    = 2r, 2r + ,rZ 2  But  = 2r + , r  Z gives extraneous roots 2 sin 5 cos–1 1    cos    4  =   n + cos = . Then 2n 2n 2 (A) 6  n  8 (B) 4  n  8 313 = 2n ± cos–1 = 2n ± sin  or, cos( – ) = where cos =  12 cos  +  ,rZ 4 1 which is not possible. hence no triangle will be possible (ii) When A < 90° and a = b sin A : In this case sin B = 1 B = 90° only one triangle is possible which is right angled at B. (iii) When A < 90° and a > b sin A : In this case sin B = b sin A a gives two such angles say B1 , B2 that B1 + B2 = 180° (iv) when A > 90° If a b , then B is also obtuse angle which is not possible. If a > b , then A > B and C will be an acute angle. So solution will exist. Note : Above case (iv) is called ambiguous case. 58