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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