化简cosθsinx-sin(x-θ)+(tanθ-2)sinx-sinθ
化简cosθsinx-sin(x-θ)+(tanθ-2)sinx-sinθ
(sinX+sinθ) .(Sinx-sinθ)=sin(x+θ).cos(x-θ)证明
化简sin^4x/sinx-cosx - (sinx+cosx)cos^2x/tan^2x-1
F(x)=sin^2θsinx+cos^2θcosx 谁能帮我化简一下?
化简[sin^2(x)]/(sinx-cosx)-(sinx+cosx)/[tan^2(x)-1]-sinx
(sin^x/sinx-cosx)-sinx+cosx/tan^2x-1
已知(2sinx+cosx)(sinx+2cosx-3)=0 求sin 2x+cos 2x/tan 2x
化简:cos^2θcotθ+sin^2θtanθ+2cosθsinθ
化简:sin^2`x/(sinx-cosx)- (sinx+cosx)/tan^2`x - 1.
求证 sin^2x/(sinx-cosx)-(sinx+cosx)/tan^2 x-1=sinx+cosx
化简sinx/1-cosx·根号下(tanθ-sinθ)/(tanθ+sinθ)(θ为第二象限角)
sinx+cosx/sinx-cosx=2 求sinx/cos^3x +cosx/sin^3x