若x=π/12.则sinx^4-cosx^4=?sinx^4+cosx^4=?
求证:tan(x-π/4)=(sinx-cosx)/(sinx+cosx)
化简f(x)=4sinx*sin^2((π+2x)/4)+(cosx+sinx)(cosx-sinx)
f(-sinx)+3f(sinx)=4sinx*cosx,
解下列方程 cos2x=cosx+sinx sin^4x-cos^4x=cosx+sinx
求证sinx-cosx=根号2sin(x-π/4)
sinx+cosx=√2sin(x+π/4)
证明1+sinx/cosx=tan(π/4+x/2)
设f(x)满足f(-sinx)+3f(sinx)=4sinx乘以cosx,(绝对值x
解方程 [1/(cosx-sinx)]^2-4[cosx/(cosx-sinx)]+2=0
2sin(x-π/4)sin(x+π/4)=(sinx-cosx)(sinx+cosx)=-cos2a
已知tan(x+(5π)/4)=2,求(sinx+cosx)/(sinx-cosx)+sin2x+cos(x)^2
已知tanX=2,求(-sinX+4cosX×sinX)/(cos²X+sinX×cosX)的值