Question

求证(1+sin4θ-cos4θ)/2tanθ=(1+sin4θ+cos4θ)/(1-(tanθ)^2)

求证(1+sin4θ-cos4θ)/2tanθ=(1+sin4θ+cos4θ)/(1-(tanθ)^2)

要详细过程~谢谢！

Answer 1

提示：左边＝（sin4θ+2sin2θsin2θ)/2tanθ
=2sin2θ(cos2θ+sin2θ)/2tanθ
=2sinθcosθ(cos2θ+sin2θ)/(sinθ/cosθ)
=2cosθcosθ(cos2θ+sin2θ)
右边仿此方法算，结果也等于这个。

Answer 2

左=(1+sin4Θ-cos4Θ)/2tanΘ =[2sin2Θ(sin2Θ+cos2Θ)]/2tanΘ=[4sinΘcosΘ(sin2Θ+cos2Θ)]/(2sinΘ/cosΘ) =2cos�0�5Θ(sin2Θ+cos2Θ)右=(1+sin4Θ+cos4Θ)/1-tan�0�5Θ =(sin4Θ+2cos2Θ)/ 1-tan�0�5=2cos2Θ(sin2Θ+cos2Θ)/[1-(sin�0�5/cos�0�5Θ)] =2cos2Θ(sin2Θ+cos2Θ)/(cos2Θ/cos�0�5Θ)=2cos�0�5Θ(sin2Θ+cos2Θ) 左=右,证毕.Ps:cos4Θ=1-2sin�0�52Θ=2cos�0�52Θ-1