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Questions  

If sin5θ=asin5θ+bsin3θ+csinθ+d, then 

a
a+b+c=0
b
a+b+c+d=0
c
5a+3b–4c=0
d
a–3c+d=0

detailed solution

Correct option is C

Sin 5θ=sin⁡(2θ+3θ)=sin⁡2θcos⁡3θ+cos⁡2θsin⁡3θ=2sin⁡θcos⁡θ×cos⁡θ4cos2⁡θ−3+1−2sin2⁡θ3sin⁡θ−4sin3⁡θ =2sin⁡θ1−sin2⁡θ1−4sin2⁡θ+1−2sin2⁡θ3sin⁡θ−4sin3⁡θ =16sin5⁡θ−20sin3⁡θ+5sin⁡θ So a=16,b=−20,c=5,d=0⇒5a+3b−4c=0

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