1+2sin(π-α)cos(4π-α)/sin²(π+α)-cos²(π-α)=tanα+1/tanα-1

已知0＜α＜π/2,且3sinα=4cosα求(sin^2α+2sinαcosα)/(3cos^2α-1)求cos^2α+sinαcosα 已知tan（π-α）=2,求sin²α-2sinαcosα-cos²α/4cos²α-3sin²α+1 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 化简：[1+2sin(α+2π)*cos(α-2π)]/[sin(α+4π)+cos(α+8π)] 求证：（1-sinα+cosα）/（1+sinα+cosα）=tan(π/4-α/2) 求证(1-2sinαcosα)/(cos²α-sin²α)=tan(π/4-α) 求证:1-2sinαcosα/cos²α-sin²α=tan（π/4-α） 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? 已知 sinα+2cos(5π/2+α)/cos(π-α)-sin(π/2-α)=-1/4 求(sinα+cosα)平方的值已知sinα+2cos(5π/2+α)/cos(π-α)-sin(π/2-α)=-1/4求：(sinα+cosα)平方的值 (1+sinα+cosα)*[sin(α/2)-cos(α/2)]/√(2+2cosα)化简 （3/2*π cosαcos[(2k+1)π]-sinαsin[(2k+1)π]为什么等于-cosα tan(π/4+α)=2,求1/(2cosαsinα+cos^2α) 化简：2cos²α-1/【2sin(π/4-α)sin²(π/4+α)】 化简：2cos²α-1/【2sin(π/4-α)sin²(π/4+α)】 若sinα=1/2,则sin(π/4-α)cos(π/4-α)= 设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π＋α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2（-sinα)(-cosα）+cosα/[1+sin² 函数y=sinα+cosα-4sinαcosα+1,且2sin^2α+sin2α/1+tanα=k,π/4 若cosα>sinα,(-π/2