Question

(2cos²x-1)/2tan(π/4-x)sin²(π/4+x)

2cos²x-1)/2tan【(π/4）-x】sin²【(π/4）+x】 求详细过程，谢谢

Answer 1

因为(π/4-x)+(π/4+x)=π/2，所以tan(π/4-x)=cot(π/4+x)
(2cos²x-1)/2tan(π/4-x)sin²(π/4+x)
=sin2x/[2cot(π/4+x)sin²(π/4+x)]
=sin2x/{[2cos(π/4+x)/sin(π/4+x)]sin²(π/4+x)}
=sin2x/[2sin(π/4+x)cos(π/4+x)]
=sin2x/sin(π/2+2x)
=sin2x/cos2x
=tan2x

追问

没教过cot，所以看不懂。

追答

cotx=1/tanx
(2cos²x-1)/2tan(π/4-x)sin²(π/4+x)
=cos2x/[2cot(π/4+x)sin²(π/4+x)]
=cos2x/{[2cos(π/4+x)/sin(π/4+x)]sin²(π/4+x)}
=cos2x/[2sin(π/4+x)cos(π/4+x)]
=cos2x/sin(π/2+2x)
=cos2x/cos2x
=1

Answer 2

(2cos²x-1)/2tan(π/4-x)sin²(π/4+x)
=cos2x/2tan(π/4-x)sin²[π/2-(π/4-x)]
=cos2x/2[sin(π/4-x)/cos(π/4-x)]cos²(π/4-x)
=cos2x/2sin(π/4-x)cos(π/4-x)
=cos2x/sin2(π/4-x)
=cos2x/sin(π/2-2x)
=cos2x/cos2x
=1