Q.
The value of sin8θ+cos8θ+sin6θcos2θ +3sin4θcos2θ+cos6θsin2θ+3sin2θcos4θ is equal to
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a
cos22θ
b
sin22θ
c
cos32θ+sin32θ
d
none of these
answer is D.
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Detailed Solution
We have, sin8θ+cos8θ+sin6θcos2θ+3sin4θcos2θ+cos6θsin2θ+3sin2θcos4θ =sin8θ+cos8θ+sin2θcos2θ(sin4θ+cos4θ+3sin2θ+3cos2θ) =(sin4θ+cos4θ)2−2sin4θcos4θ+14sin22θ{(sin2θ+cos2θ)2−12sin22θ+3} =1−sin22θ42−18sin42θ+sin22θ−18sin42θ =1+12sin22θ−316sin42θ =116(16+8sin22θ−3sin42θ)
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