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Question

If sinθ+sin2θ=1, then prove that
cos12θ+3cos10θ+3cos8θ+cos6θ1=0

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Solution

The question should be if sinθ+sin2θ=1 then prove that
cos12θ+cos6θ+3cos10θ+3cos8θ1=0

We have sinθ+sin2θ=1
sinθ=1sin2θ
sinθ=cos2θ .......... [sin²θ+cos²θ=1]

Squaring both sides, we get
sin2θ=cos4θ
1cos2θ=cos4θ ...... [sin²θ+cos²θ=1]
cos4θ+cos2θ=1

Cubing both sides,we get
(cos4θ+cos2θ)3=1

Using, (a+b)3=a3+3a2b+3ab2+b3

cos12θ+cos6θ+3cos6θ(cos4θ+cos2θ)=1

cos12θ+cos6θ+3cos10θ+3cos8θ1=0

cos12θ+3cos10θ+3cos8θ+cos6θ1=0

Hence proved.

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