Question

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Answer 1

Answer:

Given : sinθ+sin

2 θ+sin 3 θ=1

sinθ(1+sin 2 θ)=1−sin 2 θ=cos 2 θ

⇒ 1−cos 2 θ (2−cos 2 θ)=cos 2 θ

Squaring both sides, we have

(1−cos 2 θ)

(4+cos 4 θ−4cos 2 θ)=cos 4 θ

⇒cos 6 θ−4cos 4 θ+8cos 2 θ=4