设sin(π/4+α)+cos(π/4+α)=-1/5,则cos2α=
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设sin(π/4+α)+cos(π/4+α)=-1/5,则cos2α=设sin(π/4+α)+cos(π/4+α)=-1/5,则cos2α=设sin(π/4+α)+cos(π/4+α)=-1/5,则c
设sin(π/4+α)+cos(π/4+α)=-1/5,则cos2α=
设sin(π/4+α)+cos(π/4+α)=-1/5,则cos2α=
设sin(π/4+α)+cos(π/4+α)=-1/5,则cos2α=
如图.
设sin(π/4+α)+cos(π/4+α)=-1/5,则cos2α=
由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证:1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a)
设α∈(0,π/2),则cosα,sin(cosα),cos(sinα)的大小关系为什么?
利用和差角公式化简 (2)sin(π/3+α)+sin(π/3-α)(2)sin(π/3+α)+sin(π/3-α)(3)cos(π/4+α)-cos(π/4-α)(4)cos(60°+α)+cos(60°-α)(5)sin(α-β)cosβ+cos(α-β)sinβ(6)cos(α+β)cosβ+sin(α+β)sinβ
cos^2(π/4+α)=(cos(π/4)cosα-sin(π/4)sinα)² cos^2(π/4+α)是怎么等于(cos(π/4)cosα-sin(π/4)sinα)²的?
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设a为常数,解方程cos(x-π/4)=sin(2x)+a
化简sin(π/4+α)cosα-sin(π/4-α)sinα