证明恒等式,1+sin4θ-cos4θ/2tanθ=1+sin4θ+cos4θ/1-tan²θ..
来源：学生作业帮助网 编辑：六六作业网 时间：2024/05/03 12:10:16
证明恒等式,1+sin4θ-cos4θ/2tanθ=1+sin4θ+cos4θ/1-tan²θ..
证明恒等式,1+sin4θ-cos4θ/2tanθ=1+sin4θ+cos4θ/1-tan²θ..
证明恒等式,1+sin4θ-cos4θ/2tanθ=1+sin4θ+cos4θ/1-tan²θ..
首先,记得加小括号.
要证 (1+sin4θ-cos4θ)/2tanθ=(1+sin4θ+cos4θ)/1-tan²θ
只要证 (1+sin4θ-cos4θ)/(1+sin4θ+cos4θ)=2tanθ/1-tan²θ
上式右边等于tan2θ
只要证(1+sin4θ-cos4θ)/(1+sin4θ+cos4θ)=tan2θ
又 1+sin4θ-cos4θ=(1-cos4θ)+sin4θ=2sin²2θ+2sin2θcos2θ=2sin2θ(sin2θ+cos2θ)
1+sin4θ+cos4θ=(1+cos4θ)+sin4θ=2cos²2θ+2sin2θcos2θ=2cos2θ(sin2θ+cos2θ)
两式相除 =tan2θ
证毕.
[1+sin(4θ)-cos(4θ)]/[1+sin(4θ)+cos(4θ)]
=[2sin²(2θ)+2sin(2θ)cos(2θ)]/[2cos²(2θ)+2sin(2θ)cos(2θ)]
=2sin(2θ)[sin(2θ)+cos(2θ)]/2cos(2θ)[sin(2θ)+cos(2θ)]
=sin(2θ)/cos(2θ)
=tan(2θ...
全部展开
[1+sin(4θ)-cos(4θ)]/[1+sin(4θ)+cos(4θ)]
=[2sin²(2θ)+2sin(2θ)cos(2θ)]/[2cos²(2θ)+2sin(2θ)cos(2θ)]
=2sin(2θ)[sin(2θ)+cos(2θ)]/2cos(2θ)[sin(2θ)+cos(2θ)]
=sin(2θ)/cos(2θ)
=tan(2θ)=2tanθ/(1-tan²θ)
[1+sin(4θ)-cos(4θ)]/(2tanθ)=[1+sin(4θ)+cos(4θ)]/(1-tan²θ)
解题过程中有平方符号，如果手机上看不到，请到电脑上看解题过程。
收起