(1)sin(-x-5π)· cos(x-π/2)-tan(2π-x)=( )
(1)sin(-x-5π)· cos(x-π/2)-tan(2π-x)=( )
求证 tan(2π-X)sin(-2π-X)cos(6π-X)/ sin(X+3π/2)*cos(X+3π/2)=-ta
三角函数已知f(x)=(2cos^4x—2cos²x+1/2)/(2tan(π/4—x)sin²(x
化简2cos*2-1 / 2tan(π/4-x)sin*2(π/4+x)
化简:1.【(sin^2)(-X-π) *cos(π+X)cosX】/【tan(2π+X) *(cos^3 (-X-π)
已知sin x/2 -- 2cos x/2=0.(1)求tan x的值.(2)求 cos2x/(√2cos(π/4+x)
求化简数学公式哦[3sin^2(x/2)+cos^2(x/2)-4sin(x/2)cos(x/2)]/tan(π+x)化
已知sin(π+X)=-1/2.求(1)cos(2π-X) (2)tan(X-7π)
急!已知tan[(π/4)-x]=-1/3.求sin²(x+π/4)/(2cos²x+sin2x)
确定下列函数的定义域1)y=√[lg(cos x)]2) y={tan[x-π/4)]*√(sin x)}/lg(2co
若tanα=m 则sin(-5π-α)cos(3x+α)=
已知sinx+cosx/2sinx-3cosx=3,求1,tan(π-x)的值 2,5/2sin²x-3cos