Q. 34

# Mark the Correct alternative in the following:is equal toA.16 cos4 x – 12 cos2 x + 1B. 16 cos4 x + 12 cos2 x + 1C. 16 cos4 x – 12 cos2 x - 1D. 16 cos4 x + 12 cos2 x - 1

Given

Let 5x = 3x + 2x

Then

[using sin (A+B) = sin A cos B + cos A sin B]

[using the formulae :

sin 3x = 3sin x – 4 sin3x

cos 3x = 4 cos3x – 3 cos x

cos 2x = 2cos2x – 1

sin 2x = 2 sin x cos x ]

= (3 – 4 sin2x)(2cos2x -1) + (4 cos3x – 3cos x)(2cosx)

= (6cos2 x – 3 – 8 sin2 x cos2x + 4 sin2 x) + (8 cos4x - 6 cos2x)

[using sin2x + cos2 x = 1]

= – 3 – 8 (1 - cos2x) cos2x + 4 (1 - cos2x)+ 8cos4x

= – 3 – 8cos2x + 8cos4x + 4 - 4cos2x+8 cos4x

= 16 cos4x – 12 cos2x + 1

Therefore the answer is option A.

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