Question

化简[2(cosα)^2-1]/2tan[(π/4)-α]sin[(π/4)+α]

化简化简[2(cosα)^2-1]/2tan[(π/4)-α]sin[(π/4)+α]，急用啊

Answer 1

[2(cosα)^2-1]=cos2α
tan[(π/4)-α]=sin[(π/4)-α]/cos[(π/4)-α]
cos[(π/4)-α]=sin[(π/4)+α] sin[(π/4)-α]=cos[(π/4)+α]
所以
2tan[(π/4)-α]sin[(π/4)+α]=sin[(π/4)-α]
[2(cosα)^2-1]/2tan[(π/4)-α]sin[(π/4)+α]=cos2α/sin[(π/4)-α]
也可化简为cotα*cos[（π/4)-α]
cotα*sin[(π/4)+α]
用到的三角函数公式：
cos(α+β)=cosαcosβ-sinαsinβ
cos(α-β)=cosαcosβ+sinαsinβ
sin(α+β)=sinαcosβ+cosαsinβ
sin(α-β)=sinαcosβ -cosαsinβ 　
sin2A=2sinA·cosA 　
1.Cos2a=Cos^2(a)-Sin^2(a) 　　
2.Cos2a=1-2Sin^2(a) 　　
3.Cos2a=2Cos^2(a)-1
tanα=sinα/cosα