Question

三角函数证明(sinα+sinθ)*(sinα-sinθ)=sin(α+θ)*sin(α-θ)

求证(sinα+sinθ)*(sinα-sinθ)=sin(α+θ)*sin(α-θ)

Answer 1

(sinα+sinθ)*(sinα-sinθ)=sin(α+θ)*sin(α-θ)
sin²桐念α-sin²θ=（sinαcosθ+sinθcosα）*（sinαcosθ-sinθcosα）
sin²α-sin²θ=（sinαcosθ）²-（sinθcosα）²
sin²α-sin²θ=sin²α*cos²θ-sin²θcos²α
sin²α（局旅困1-cos²θ）=sin²θ（1-cos²α）
sin²αsin²θ=sin²镇缺θsin²α

Answer 2

证明：戚缓
(sinα+sinθ)(sinα-sinθ)
=sin²α-sin²θ
=[(1-cos2α)-(1-cos2θ)]/2
=(cos2θ-cos2α)/庆伍2
={cos[(α+θ)-(α-θ)]-cos[(α+θ)+(α-θ)]}/2
={[cos(α+θ)cos(α-θ)+sin(α+θ)sin(α-θ)]-[cos(α+θ)cos(α-θ)-sin(α+θ)sin(α-θ)]}/2
=[2sin(α+θ)sin(α-θ)]/2
=sin(α+θ)sin(α-θ)
证誉仔或毕