[2(cosα)^2-1]/2tan(π/4-α)*[sin(π/4+α)]^2化简
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/03 00:49:36
[2(cosα)^2-1]/2tan(π/4-α)*[sin(π/4+α)]^2化简
注释:此题有可能是书写错误,可能是后面的括号写掉了.
我想原题应该是[2(cosα)^2-1]/[2tan(π/4-α)*[sin(π/4+α)]^2]化简.
它的化简结果是1.解题过程如下:
[2(cosα)^2-1]/[2tan(π/4-α)*[sin(π/4+α)]^2]
=cos(2α)/2*(cos(π/4-α)/sin(π/4-α))*(2/(1-cos(π/2+2α))
=cos(2α)/2*(1+cos(π/2-2α)/sin(π/2-2α))*(2/(1+sin2α))
=(cos(2α)/(1+sin2α))*((1+sin2α)/cos(2α))
=1
我想原题应该是[2(cosα)^2-1]/[2tan(π/4-α)*[sin(π/4+α)]^2]化简.
它的化简结果是1.解题过程如下:
[2(cosα)^2-1]/[2tan(π/4-α)*[sin(π/4+α)]^2]
=cos(2α)/2*(cos(π/4-α)/sin(π/4-α))*(2/(1-cos(π/2+2α))
=cos(2α)/2*(1+cos(π/2-2α)/sin(π/2-2α))*(2/(1+sin2α))
=(cos(2α)/(1+sin2α))*((1+sin2α)/cos(2α))
=1
求证:(1-sinα+cosα)/(1+sinα+cosα)=tan(π/4-α/2)
求证:1-2sinαcosα/cos²α-sin²α=tan(π/4-α)
试证明:1+2sinαcosα/cos平方α-sin平方α=tan(π/4-α)
求证(1-2sinαcosα)/(cos²α-sin²α)=tan(π/4-α)
已知tan(α+π/4)=3,求cosα+2sinα/cosα-sinα的值
化简:(2cos^2α-1)/[2tan(π/4-α)sin^2(π/4+α)]
已知tan(π-α)=2,求sinα-2sinαcosα-cosα/4cosα-3sinα的值
化简(2cosα-1)/(2tan(π/4-α)*sin(π/4+α))
化简(2cos²α-1)/2tan(π/4+α)sin²(π/4-α)
化简(2cos²α-1)/[2tan(π/4-α)sin²(π/4+α)]
化简:[sin(π+α)*cos(π-α)*tan(π-α)]/[cos(π/2+α)*tan(3π/2-α)*tan(
证明[2-2sin(α+3π/4)cos(α+π/4)]/(cos^4α-sin^4α)=(1+tanα)/(1-tan