化简cos2x/[2cot(π/4+x)cos平方(π/4-x)]
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/13 21:44:29
化简cos2x/[2cot(π/4+x)cos平方(π/4-x)]
因为cos²(π/4-x)= cos²[π/2- (π/4+x)]
= sin²(π/4+x)
所以2cot(π/4+x)cos²(π/4-x)
=2cot(π/4+x) sin²(π/4+x)
=2cos(π/4+x) sin(π/4+x)
= sin[2(π/4+x)]
= sin(π/2+2x)
= cos2x
∴cos2x/[2cot(π/4+x)cos²(π/4-x)]=1.
= sin²(π/4+x)
所以2cot(π/4+x)cos²(π/4-x)
=2cot(π/4+x) sin²(π/4+x)
=2cos(π/4+x) sin(π/4+x)
= sin[2(π/4+x)]
= sin(π/2+2x)
= cos2x
∴cos2x/[2cot(π/4+x)cos²(π/4-x)]=1.
已知4sin^2x-cos^2x+3cosx=0求(cos2x-cos^2x)/(1-cot^2x)
化简lg[cos(2π-x)cot(3π/2-x)+1-2xin^2x/2]+lg[根号2cos(x-π/4)]-lg[
cos[2(π-x)]=-cos2x对不对?
化简cos(π/2-x)cos(π/2+x)cot(π-x)/sin(3π/2+x)cos(3π+x)tan(π+x)
求证:sin(π/4+x)/sin(π/4-x)+cos(π/4+x)/(π/4-x)=2/cos2x
求下列不定积分:1、(cot)^2•xdx 2、cos2x/(cos^2xsin^2x)dx
化简:2cos2x+2sin^2 x+cos(-x)分之sin2x+sin(π-x)=___________
已知函数f(x)=(4cos^4x-2cos2x-1)/[sin(π/4+x)·sin^2(π/4-x)]化简
已知函数f(x)=(4cos^4x-2cos2x-1)/[sin(π/4+x)·sin(π/4-x)]化简
已知tanx=-4/3,试化简(2+cos2x)/[(根号2)·cos(x-π/4)-sin2x]
若cos(π/2+x)=4/5,则cos2x=
cos2x/根号2cos(x+π/4)=1/5,0