用cosx表示sin^4x-sin^2x+cos^2x

用cos x表示sin^4 x -sin^2 x +cos^2 x 化简sin^4x/sinx-cosx - (sinx+cosx)cos^2x/tan^2x-1 若|sinx|>|cosx|,则sin^2 x>cos^2 x 求化简(sinx+tanx)/cos^2x+sin^2x+cosx sin(x+1/2π)=cosx?还是cos-x 已知函数f(x)=2(sin^4 x+cos^4 x)+m(sin^x+cosx)^4在0= f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx 证明成立：[cos(3x)-sin(3x)]/(cosx+sinx)=1-2sin(2x). 求证:Sin^2 x / (sinx-cosx) - (sin x + cos x)/(ta 已知函数f(x)=cos^4 x -2sinx *cosx-sin^4 x sin(-x)=sin(x),cos(-x)=-cosx对吗? sinx+cosx/sinx-cosx=2 求sinx/cos^3x +cosx/sin^3x