Trigonometry Solved Problems

TRIGONOMETRY SOLVED PROBLEMS Ex.1 Find the value of the expression - = 2 1 cos y sin y sin y 1 – + – sin y 1  cos y 1  cos y Sol. 1 cos y sin y sin2 y 1 – + – 1  cos y sin y 1  cos y 2 = cos y  cos 2 y + 0 = cos y 1  cos y = Sol. Sol. = If cosec  –sin =m and sec–cos = n then find (m2n)2/3 + (n2m)2/3 cosec  – sin  = m 1 cos 2  m = – sin  = ...(i) sin sin  n = 1 sin 2  – cos  = cos  cos  ...(ii) cos 2  sin 2  . = sin  cos  sin  cos  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 m × n = Ex.3   Find the value of  1  cos  8 5    1  cos   8     1  cos   8 Sol. 7    1  cos   8 3    1  cos   8   3     1  cos      8   = 3    1  cos   8      1  cos        8    3    1  cos   1  cos     8 8  3      1  cos   1  cos     8 8  2   2 3   =  1  cos   1  cos 8 8  = 1  2  1  cos    2  1  cos 3       4  4  4  = 1  1  cos    1  cos 3       4  4  4  Ex.5  1  1  1 1  1 1  1    1   =   = 2 2 2  4  8  Find the value of sin 20° sin 40° sin 60° sin 80° sin 20° sin 40° sin 60° sin 80° 2 1  cos y  sin y 1  cos y  sin y + 1  cos y sin y (1  cos y ) = Ex.2 2 Ex.4 1 4 3 sin 20° sin (60° – 20°) sin (60° + 20°) 2 3 sin 20° (sin2 60° – sin2 20°) 2 = 3 3 sin 20° ( – sin2 20°) 4 2 = 3 (3 sin 20° – 4 sin3 20°) 8 = 3 sin 60° 8 = 3 . 8 Find cos4 3 3 = 16 2  3 5 + cos4 + cos4 8 8 8 7 8  3 = cos4 +cos4 + cos4 8 8 5 7 + cos4 8 8  3  = cos4 + cos4 + cos4 8 8 3  + cos4 8 8 + cos4 Sol. 3   4   cos 4  = 2  cos 8 8  2 2 1   2  2 3  = 2  2 cos 8    2 cos 8        1   3   = 2 1  cos 4   1  cos 4      2 2 2 1  1   1   1       1 –     = 2  2 2       1 3 2  1 = = 2 2 2   TRIGONOMETRY Ex.6 Sol. 3 , then prove that 2 cos 2A+cos2B+cos2C=1–4sinAsinBsinC cos 2A + cos 2B + cos 2C = 2cos (A + B) cos (A – B) + cos 2C  3   C cos (A – B) + cos 2C = 2 cos   2 If A + B + C= 3 2 = –2 sin C cos (A – B) + 1 – 2 sin2 C = 1 – 2sin C [cos (A – B) + sin C) Ex.9 Sol.  A + B + C =   3  = 1 – 2sinC cos ( A  B)  sin 2  A  B     = 1 – 2 sinC [cos (A – B) – cos (A + B)] = 1 – 4 sin A sin B sin C Ex.7 Sol. In any triangle ABC, sin A–cosB=cosC, then find angle B We have, sin A – cos B = cos C sin A = cos B + cos C 2 sin A A cos 2 2 2 sin A A cos 2 2  B  C B  C  cos   = 2cos   2  2     A B  C = 2 cos   cos    2   2  A A A + B + C =  2 sin cos 2 2 B  C   A  = 2 sin cos  2  2  A BC = cos 2 2 or A = B – C But A + B + C =  Therefore 2B =   B = /2 If sin 3 = sin, then find the general value of  sin 3 = sin or, 3 = m + (–1)m For (m) even i.e. m = 2n, 2n then  = = n 2 and for (m) odd i.e. m = (2n + 1)  or,  = (2n +1) 4 Ex.10 Find the number of solutions of equation, sin 5x cos 3x = sin 6x cos 2x, in the interval [0, ] Sol. The given equation can be written as 1 1 (sin 8x + sin 2x) = (sin 8x + sin 4x) 2 2 or, sin 2x – sin 4x  – 2 sin x cos 3x = 0 Hence sin x = 0 or cos 3x = 0. That is, x = n (n  I), or  3x = k + 2 (k  I). Therefore, since x  [0,  the given equation is satisfied   5 if x = 0, , , or . 6 2 6 Ex.11 Find the general solution of     tan  sin  = cot  cos  2 2 Sol.     We have, tan  sin  = cot  cos  2 2  cos Ex.8 Sol. Find tan 9° – tan27° – tan 63° + tan 81° tan 9° + tan 81° – (tan 27° + tan 63°) (tan 9° + cot 9°) – (tan 27° + cot 27°)  sin 9º cos 9º   sin 27º cos 27º  =  cos 9º  sin 9º  –  cos 27º  sin 27º      1 1 = sin 9º cos 9º – cos 27 º sin 27 º 2 2 2 2 = –  =  – sin 36 º sin 54 sin 18 sin 18 = = 2 4 5 1 – 16 = 4 4  5  1 5  1   = 8  5 1  ( 5  1)( 5  1)  24              tan  sin  = tan   cos  2 2 2   sin  = r  + – 2 2 r  Z sin  + cos  = (2r + 1 1 sin  + cos  2 2 r  Z  cos , 2 1), r  Z 2r  1 = , 2 2r  1   cos     = , r  Z 4 2 1 1   cos     = or – 4 2 2   = 2r ± , r  Z 4 4   q = 2r + + , r  Z 4 4   = 2r, 2r + , r  Z 2  – TRIGONOMETRY  , r  Z gives extraneous 2 roots as it does not satisfy the given equation. Therefore  = 2r, r  Z But  = 2r + Ex.12 Find the general solution of the equation sec 4 – sec 2 = 2 Sol. Given equation is, sec 4 – sec 2 = 2 or, 1 1  = 2, cos 4 cos 2 cos 4  0, cos 2  0 cos 2 – cos 4 = 2 cos 4 cos 2 cos 2 – cos 4 = cos 6 + cos 2 cos 6 + cos 4 = 0 2 cos 5 cos  = 0 either cos 5 = 0 or cos  = 0 cos 5 = 0, then 5 = (2n + 1) /2 or,  = (2n + 1) /10 where n  I. and if cos  = 0, then  = (2n + 1) /2, where n  I. obviously for   = (2n + 1) /2 and = (2n + 1) /10, cos 2 or cos 4 are not zero Hence  = (2n + 1) /2, (2n + 1) /10 are the general solutions of the given equation. or, or, or, or  If Ex.13 A tower subtends an angle of 30º at a point on the same level as its foot, and at a second point h m above the first, the depression of the foot of tower is 60º .Find the height of the tower Sol. Let OP be the tower of height x., A the point on the same level as the foot O of the tower and B be the point h m above A (see Ex.14 A tower subtends an angle of 30º at a point on the same level as the foot of the tower . At a second point, h metre above first, the depression of the foot of the tower is 60º, find the horizontal distance of the tower from the points Sol. Let the tower OP subtend an angle of 30º at A.a point on the same level as the foot of the tower. From a point B at the height h, vertically above A, from where the angle of depression of foot O of the tower is 60º i.e.,  AOP = 60º. From  OAP , OA = h cot 60º . Also from  OAP , OA = OP cot 30º .  OA = h cot 60º = OP cot 30º  OP = h/3.  OA = (h/3) cot 30º, which is given in (B) Ex.15 The top of a hill observed from the top and bottom of a building of height h is at angles of elevation p and q respectively. Fnd the height of the hill Sol. Let AD be the building of height h and BP be the hill. Then Fig.). Then  AOB = 60º and  PAO = 30º . From right-angled triangle AOP, we have OA = x cot 30º and from right-angled triangle OAB, we have OA = h cot 60º Therefore, from (1) and (2), we get x cot 30º = h cot 60º B   P h 30º x 60º O 3x = 30º 1 3 h, A 1 x = h 3 hx x and tan p = y y hx tan q = x cot p x cot p = (h + x) cot q h cot q x = cot p  cot q h cot q h + x = h + cot p  cot q h cot p = cot p  cot q tan q = 