证明(sina+sinθ)*(sina-sinθ)=sin(a+θ)*sin(a-θ)

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证明(sina+sinθ)*(sina-sinθ)=sin(a+θ)*sin(a-θ)

sin(a+θ)*sin(a-θ)
=(sinacosθ+cosasinθ)*(sinacosθ-cosasinθ)
=sin²acos²θ-cos²asin²θ
=sin²a(1-sin²θ)-(1-sin²a)sin²θ
=sin²a-sin²asin²θ-sin²θ+sin²asin²θ
=sin²a-sin²θ
=(sina+sinθ)*(sina-sinθ)
所以
(sina+sinθ)*(sina-sinθ)=sin(a+θ)*sin(a-θ)

两边都打开,肯定能约掉