PROVING PROBLEMS IN TRIGONOMETRY

Problem 1 :

Prove that

cos π/15 cos 2π/15 cos 3π/15 cos 4π/15 cos 5π/15 cos 6π/15 cos 7π/15  =  1/128

Solution :

  =  cos π/15 cos 2π/15 cos 3π/15 cos 4π/15 cos 5π/15 cos 6π/15 cos 7π/15

π/15  =  180/15  =  12

2π/15  =  24 , 3π/ 15  =  36, 4π/ 15  =  48

5π/ 15  =  60, 6π/ 15  =  72, 7π/ 15  =  84

  =  cos 12 cos 24 cos 36 cos 48 cos 60 cos 72 cos 84

To solve this problem, we use the property given below.

cos (60 - A) cos A cos (60 + A)  =  1/4 (cos 3A)

  =  (1/2) cos 48 cos 12 cos 72 cos 24 cos 36 cos 84

48  =  60 - 12 and 72  =  60 + 12

36  =  60 - 24 and 84  =  60 + 24

  =  (1/2) cos(60-12) cos12 cos(60+12)cos 36 cos 24 cos 84

  =  (1/2) cos(60-12) cos12 cos(60+12)cos(60-24) cos 24 cos (60+24)

Instead of first 3 terms, we may use the formula (1/4) cos 3A

  =  (1/2) (1/4) cos 36 (1/4) cos 72

  =  (1/16) (√5 + 1)/4 (√5 - 1)/4

  =  (1/256) (5 - 1)

  =  (4/256) 

  =  1/128

Problem 2 :

Prove that

(sin8x cosx-sin6x cos3x)/(cos2x cosx-sin3x sin4x) = tan2x 

Solution :

  =  (sin8x cosx - sin6x cos3x)/(cos2x cosx - sin3x sin4x)

sin8x cosx  =  (1/2) (2 sin8x cosx)

  =  (1/2) [sin 9x + sin 7x]  ---------(1)

sin6x cos3x  =  (1/2) (2 sin6x cos3x)

  =  (1/2) [sin 9x + sin 3x]  ---------(2)

cos2x cosx  =  (1/2) (2 cos2x cosx)

  =  (1/2) [cos 3x + cos x]  ---------(3)

sin3x sin4x  =  (1/2) (2 sin3x sin4x)

  =  (1/2) [cos x - cos 7x]  ---------(4)

(1) - (2) 

  =  (1/2) [sin 9x + sin 7x] - (1/2) [sin 9x + sin 3x]

  =  (1/2)[sin 9x + sin 7x - sin 9x - sin 3x]

  =  (1/2)[sin 7x  - sin 3x]  -------(A)

(3) - (4)

  =  (1/2) [cos 3x + cos x] - (1/2) [cos x - cos 7x]

  =  (1/2) [cos 3x + cos x - cos x + cos 7x]

  =  (1/2) [cos 3x + cos 7x] -------(B)

(A)/(B)

  =  [sin 7x  - sin 3x]/[cos 3x + cos 7x]

Using the formula sin C - sin D and cos C + cos D, we get

  =  2 cos5x sin 2x / 2 cos 5x cos 2x

  =  tan 2x

Hence proved.

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