(sinx+cosx)^2+2sin^2(π/4-x)
(sin^x/sinx-cosx)-sinx+cosx/tan^2x-1
求证sinx-cosx=根号2sin(x-π/4)
sinx+cosx=√2sin(x+π/4)
2sin(x-π/4)sin(x+π/4)=(sinx-cosx)(sinx+cosx)=-cos2a
利用cosx=sin(π/2-x),sin'x=cosx,证明(cosx)'=-sinx
化简f(x)=4sinx*sin^2((π+2x)/4)+(cosx+sinx)(cosx-sinx)
化简sin^4x/sinx-cosx - (sinx+cosx)cos^2x/tan^2x-1
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
∫sinx/(cosx-sin^2x)dx
已知(sinx+cosx)/(sinx-cosx)=3,求tanx,2sin²x+(sinx-cosx)&su
证明x∈(0,π/2),cos(cosx)>sin(sinx)