化简cos2x/【sin(x+π/4)sin(x-π/4)】=?
已知函数f(x)=cos2x/[sin(π/4-x)]
化简函数f(x)=[(1+cos2x)²-2cos2x-1]/[sin(π/4+x)*sin(π/4-x)]
已知函数f(x)=(4cos^4x-2cos2x-1)/[sin(π/4+x)·sin^2(π/4-x)]化简
已知函数f(x)=(4cos^4x-2cos2x-1)/[sin(π/4+x)·sin(π/4-x)]化简
已知函数f(x)=sin(π-x)sin(π2-x)+cos2x
已知函数f(x)=2sin²(π/4+x)-根号3cos2x
f(x)=((1+cos2x)^2-2cos2x-1)/(sin(π/4+x)sin(π/4-x))
已知函数f(x)=2sin^2(π/4+x)-根号3cos2x
f(x)=4sin^2(π/4+x)-2根号3cos2x-1化简!
1.y=cos^4x+sin^4x 求周期 2.y=(sin2x+sin(2x+π/3))/( cos2x+cos(2x
若(cos2x)/[sin(x-π/4)]=(-根号2)/2,则cosx+sinx=?
求证:sin(π/4+x)/sin(π/4-x)+cos(π/4+x)/(π/4-x)=2/cos2x