Question

求证 【sinθ(1+sinθ)+cosθ(1+cosθ)】×【sinθ(1-sinθ)+cosθ(1-cosθ)】=sin2θ

要过程喔~~~

Answer 1

【sinθ(1+sinθ)+cosθ(1+cosθ)】×【sinθ(1-sinθ)+cosθ(1-cosθ)】
=【sinθ+sin^2θ+cosθ+cos^2θ】×【sinθ-sin^2θ+cosθ-cos^2θ】
=【sinθ+cosθ+(sin^2θ+cos^2θ)】×【sinθ+cosθ-(sin^2θ+cos^2θ)】
=【sinθ+cosθ+1】×【sinθ+cosθ-1】
=(sinθ+cosθ)^2 -1^2
=sin^2θ+cos^2θ+2sinθcosθ-1
=1+sin2θ-1
=sin2θ

Answer 2

色他用X代替
(sinX+sinX^2+cosX+cosX^2)(sinX-sinX^2+cosX-cosX^2)
=(sinX+cosX+1)(sinX+cosX-1)
=sinX^2+COSX^2+2sinXcosX-1
=2sinXcosX
=sin2X