化简2sin(π-x)cosx+2cos²x
sin(x+1/2π)=cosx?还是cos-x
化简:1.【(sin^2)(-X-π) *cos(π+X)cosX】/【tan(2π+X) *(cos^3 (-X-π)
化简sin^4x/sinx-cosx - (sinx+cosx)cos^2x/tan^2x-1
证明x∈(0,π/2),cos(cosx)>sin(sinx)
化简f(x)=sin(x/2+π/12)乘以cos(x/2+π/12)-cosx/2乘以cosx/2
f(x)=2cos*sin(x+π/3)-^3sin^2x+sinx*cosx
化简1+sin²x-cos²x-sinx/2sinxcosx-cosx
已知函数fx=[cosx+cos(π/2-x)][cosx+sin(π+x)]
已知(sinx+cosx)/(cosx-sinx)=2,则2sin²x-sinxcosx+cos²x
若|sinx|>|cosx|,则sin^2 x>cos^2 x
求化简(sinx+tanx)/cos^2x+sin^2x+cosx
设x∈(0,π/2),则函数(sin²x+1/sin²)(cosx²+1/cos²