化简(2cosx^4-2cosx^2+1/2)/(2tan(π/4-x)sin(π/4+x)^2)
答案:2 悬赏:40
解决时间 2021-04-07 01:15
- 提问者网友:冷场帝
- 2021-04-06 17:05
过程谢谢
最佳答案
- 二级知识专家网友:如果这是命
- 2021-04-06 17:23
=2(cos(x)*cos(x)-1/2)^2/(2tan(π/4-x)cos(π/4-x)^2)
=(1/2)*cos^2(2x)/[2sin(π/4-x)*cos(π/4-x)]
=(1/2)*cos^2(2x)/sin(π/2-2x)
=(1/2)cos(2x)
=(1/2)*cos^2(2x)/[2sin(π/4-x)*cos(π/4-x)]
=(1/2)*cos^2(2x)/sin(π/2-2x)
=(1/2)cos(2x)
全部回答
- 1楼网友:承载所有颓废
- 2021-04-06 18:41
题目应该是这样的吧:(2cos^2 x-1)/[2tan(tt/4-x)sin^2(tt/4+x)]
解:(2cos^2 x-1)/[2tan(tt/4-x)sin^2(tt/4+x)]
=(1+cos2x-1)/[2tan(tt/4-x)sin^2(tt/4+x)]
=cos2x/[2tan(tt/4-x)sin^2(tt/4+x)]
2tan(tt/4-x)sin^2(tt/4+x)=[2(sin(π/2-2x))/(cos(π/2-2x)+1)]x[(1-cos(2x+π/2))/2]
=[2(cos2x/(sin2x+1)]x[(1+sin2x)/2]
=cos2x
所以,(2cos^2 x-1)/[2tan(tt/4-x)sin^2(tt/4+x)]
=(1+cos2x-1)/[2tan(tt/4-x)sin^2(tt/4+x)]
=cos2x/[2tan(tt/4-x)sin^2(tt/4+x)]
=cos2x/cos2x
=1
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