(sinx+cosx)^2+2cos^2x-2怎么能变到1+sin2x+1+cos2x-2
Cosx+sinx分之1+sin2x+cos2x=2cos 证明
sinx+2cosx=0.求cos2x-sin2x/1+cos^x的值
求证:(1-2sinx×cosx)/cos2x-sin2x=(cos2x-sin2x)/(1+2sinx×cosx)
设(2cosx-sinx)(sinx+cos2x+3)=0,则2cos^2x+sin2x/1+tanx=?
已知sinx=-1/2cosx,cosx-sinx/cosx+sinx+sin2x+cos2x的值
化简sin2x*tanx cos2x*1/tanx 2sinx*cosx
sin3x+cosx-cos3x-sinx/2(sin2x+cos2x)
f(x)=(sinx)^4+(cosx)^2+1/4*sin2x*cos2x,则f(x)
已知x∈(0,π),且sinx+cosx=1/2 求sin2x+cos2x,sinx-cosx的值
sinx+cosx=1/5,x属于(π/2,3π/4),求sinxcosx,sin2x,cos2x,sinx,cosx
已知2sinx=cosx,求cos2x+cos2x+1/cos^2x的值.
sinx-2cosx=0.求sin2x-cos2x/1+sin^2x