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Q.

cosθ+cos2θ=1, a sin12θ+bsin10θ+csin8θ+dsin6θ=1⇒b+ca+d=

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a

0

b

1

c

2

d

3

answer is D.

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Detailed Solution

cosθ=1−cos2θ⇒cosθ=sin2θ⇒cos2θ=sin4θ⇒(sin4θ+sin2θ)=1cubing on both sides sin2x+sin4x3=1⇒sin12x+3sin10x+3sin8x+sin6x=1a=1,b=3,c=3,d=1
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