化简sin(π/4-α)+cos(π/4+α)

化简sin(π/4-α)+cos(π/4+α) 化简【cos(π/4+α)-sin(π/4+α)】/【cos(π/4-α)+sin(π/4-α)】.如题 化简:sin(π/4-α)cos(π/3-α)-cos(π/6+α)sin(π/4+α)= 化简：[1+2sin(α+2π)*cos(α-2π)]/[sin(α+4π)+cos(α+8π)] 已知 sinα+2cos(5π/2+α)/cos(π-α)-sin(π/2-α)=-1/4 求(sinα+cosα)平方 已知α∈（0,π/2),且2sinα-sinαcosα-3cosα=0.求[sin(α+π/4)]/[sin2α+cos 已知tan（π-α）=2,求sinα-2sinαcosα-cosα／4cosα-3sinα的值 已知sin a=2cos a 计算⑴（sinα+2cosα)/(5cosα-sinα) ⑵tan(α+π/4) 已知(sinα-cosα)/(sinα+cosα)=1/3,则cos^4(π/3+α)-cos^4(π/6-α)的值为 α属于（0,π/2）且2sin²α-sinαcosα-3cos²α=0求sin(α+π/4)/sin 1.化简sin²α+cosαcos(π/3+α)-sin²(π/6-α) 求证:1-2sinαcosα/cos²α-sin²α=tan（π/4-α）