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The value of sin8θ+cos8θ+sin6θcos2θ +3sin4θcos2θ+cos6θsin2θ+3sin2θcos4θ is equal to

a
cos2⁡2θ
b
sin2⁡2θ
c
cos3⁡2θ+sin3⁡2θ
d
none of these

detailed solution

Correct option is D

We have, sin8⁡θ+cos8⁡θ+sin6⁡θcos2⁡θ+3sin4⁡θcos2⁡θ+cos6⁡θsin2⁡θ+3sin2⁡θcos4⁡θ =sin8⁡θ+cos8⁡θ+sin2⁡θcos2⁡θ(sin4⁡θ+cos4⁡θ+3sin2⁡θ+3cos2⁡θ) =(sin4⁡θ+cos4⁡θ)2−2sin4⁡θcos4⁡θ+14sin2⁡2θ{(sin2⁡θ+cos2⁡θ)2−12sin2⁡2θ+3} =1−sin2⁡2θ42−18sin4⁡2θ+sin2⁡2θ−18sin4⁡2θ =1+12sin2⁡2θ−316sin4⁡2θ =116(16+8sin2⁡2θ−3sin4⁡2θ)

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