1)诺sin(π/6-x)=1/3,则cos(2π/3+2x)等于
1)诺sin(π/6-x)=1/3,则cos(2π/3+2x)等于
已知f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+cos平方x.
化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1
化简(1)√3sin x+cos x (2)√2(sin x-cos x) (3)√2cos x-√6sin x
已知sin X+cos X=m,|m|小于等于根号2且|m|不等于1,求sin X^3+cos X^3,sin X^4+
求证 tan(2π-X)sin(-2π-X)cos(6π-X)/ sin(X+3π/2)*cos(X+3π/2)=-ta
已知函数 y=sin(2x-(π/3))+cos(2x-(π/6))+2cos²x-1,x∈R
设(2cosx-sinx)(sin^2x+2cos^2x)=0,则(2cos^2x+sin^2x)/(1+tanx)等于
若sin(兀/3一x)=1/4,则cos(兀/3+2x)等于
已知sin(x-4分之π)等于3分之1,则cos(4分之π+x)等于?
sin(2x+ π/3)乘cos(2x+π/3)等于
已知sin(x-3π/4)cos(x-π/4)=-1/4,则cosx等于