利用cosx=sin(π/2-x),sin'x=cosx,证明(cosx)'=-sinx
利用cosx=sin(π/2-x),sin'x=cosx,证明(cosx)'=-sinx
求证 sin^2x/(sinx-cosx)-(sinx+cosx)/tan^2 x-1=sinx+cosx
sinx+cosx/sinx-cosx=2 求sinx/cos^3x +cosx/sin^3x
证明cosx(cosx-cosy)+sinx(sinx-siny)=2sin(x-y)/2
已知(sinx+cosx)/(sinx-cosx)=3,求tanx,2sin²x+(sinx-cosx)&su
如何证明tan x/2 =sin/(1+cosx)=(1-cosx)/sinx
证明成立:[cos(3x)-sin(3x)]/(cosx+sinx)=1-2sin(2x).
求证sinx-cosx=根号2sin(x-π/4)
sinx+cosx=√2sin(x+π/4)
已知sinx=2/3,求(cosx-sinx/cosx+sin)+(cosx+sin/cosx-sinx)的值.
2sin(x-π/4)sin(x+π/4)=(sinx-cosx)(sinx+cosx)=-cos2a
已知函数f(x)=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx