急急急!一道数学题,要详细过程:化简,cos(π/6-x)+2sin(x-π/3)-√3*(cos 2π/3-x)
急急急!一道数学题,要详细过程:化简,cos(π/6-x)+2sin(x-π/3)-√3*(cos 2π/3-x)
化简sin(3π-x)cos(x-3/2)cos(4π-x)/tan(x-5π)cos(π/2+x)sin(x-5/2π
已知函数f(x)=cos(2x-π/3)+sin^2 x-cos^2 x
已知函数f(x)=cos(2x-π\3)+sin²x-cos²x
化简(1)√3sin x+cos x (2)√2(sin x-cos x) (3)√2cos x-√6sin x
急.设函数f(x)=cos(2x+π/3)+sin^2 X
化简cos(π/2-x)cos(π/2+x)cot(π-x)/sin(3π/2+x)cos(3π+x)tan(π+x)
化简:1.【(sin^2)(-X-π) *cos(π+X)cosX】/【tan(2π+X) *(cos^3 (-X-π)
化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1
已知函数f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
已知f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+cos平方x.
化简:1、sin(x-π/3)-cos(x+π/6)+√3cosx=?2、已知,sinα+sinβ=√2/2,求cosα