求证(1+tan4θ-cos4θ)÷(2tanθ)=(1+tan4θ+cos4θ)÷(1-tan^2 θ)
网友回答
你的题目有误,应该是sin4θ
∵ (1+sin4θ-cos4θ)/(1+sin4θ+cos4θ)
=(2sin²2θ+2sin2θcos2θ)/(2cos²2θ+2sin2θcos2θ)
=2sin2θ(sin2θ+cos2θ)/[2cos2θ(cos2θ+sin2θ)]
=tan2θ
=2tanθ/(1-tan²θ)
∴ (1+sin4θ-cos4θ)÷(2tanθ)=(1+sin4θ+cos4θ)÷(1-tan^2 θ)