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Questions  

 sin7θ+6sin5θ+17sin3θ+12sinθsin6θ+5sin4θ+12sin2θ is equal to

a
2 cosθ
b
cosθ
c
2 sinθ
d
sinθ

detailed solution

Correct option is A

We have,sin⁡7θ+6sin⁡5θ+17sin⁡3θ+12sin⁡θsin⁡6θ+5sin⁡4θ+12sin⁡2θ=  (sin⁡7θ+sin⁡5θ)+5(sin⁡5θ+sin⁡3θ)+12(sin⁡3θ+sin⁡θ)sin⁡6θ+5sin⁡4θ+12sin⁡2θ=  2sin⁡6θcos⁡θ+10sin⁡4θcos⁡θ+24sin⁡2θcos⁡θsin⁡6θ+5sin⁡4θ+12sin⁡2θ=  2cos⁡θ(sin⁡6θ+5sin⁡4θ+12sin⁡2θ)sin⁡6θ+5sin⁡4θ+12sin⁡2θ=  2cos⁡θ

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