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cos12.cos24.cos36.cos48.cos60.cos72.cos84= ? Option 1/16 ,1/32, 1/64, 1/128

cos12.cos24.cos36.cos48.cos60.cos72.cos84= ?   Option  1/16  ,1/32, 1/64, 1/128

Grade:12th pass

1 Answers

Arun
25750 Points
4 years ago
Recall that sin2A = 2sinA cosA

cos12 cos24 cos36 cos48 cos72 cos84

let x = 12, then the expression becomes
cosx cos2x cos3x cos4x cos6x cos8x
= (sinx/sinx) cosx cos2x cos3x cos4x cos6x cos8x
= (1/2^4) (2^3/sinx) (2sinx cosx) cos2x cos3x cos4x cos6x cos8x
= (1/2^4) (2^2/sinx) (2sin2x cos2x) cos3x cos4x cos6x cos8x
= (1/2^4) (2^1/sinx) (2sin4x cos4x) cos3x cos6x cos8x
= (1/2^4) (1/sinx) (2sin8x cos8x) cos3x cos6x
= (1/2^4) (1/sinx) (sin16x) cos3x cos6x
= (1/2^4) (1/sinx) (sin16x) * (1/2^2) (2/sin3x) (2sin3x cos3x) cos6x
= (1/2^4) (1/sinx) (sin16x) * (1/2^2) (1/sin3x) (2sin6x cos6x)
= (1/2^4) (1/sinx) (sin16x) * (1/2^2) (1/sin3x) (sin12x)
= (1/2^6) (sin16x / sinx) * (sin12x / sin3x)

Knowing that x = 12, then 16x = 192 = 180 + 12
sin16x = sin192 = sin(180 + 12) = sin12 = sinx

Similarly, 3x = 36, 12x = 144
sin12x = sin144 = sin(180 - 36) = sin36 = sin3x

= (1/2^6) (sinx/sinx) * (sinx/sinx)
= (1/64) * 1 * 1
= 1/64

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