已知(x)=[sin(π-x)cos(2π-x)tan(-x+π)]/[cos(-π/2+x)]
已知(x)=[sin(π-x)cos(2π-x)tan(-x+π)]/[cos(-π/2+x)]
已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x
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
已知f(x)=(sin(π-x)*cos(2π-x))/cos(-π-x)*tan(π-x),则f(-31π/3)的值为
三角函数已知f(x)=(2cos^4x—2cos²x+1/2)/(2tan(π/4—x)sin²(x
已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?
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sin x/2 -- 2cos x/2=0.(1)求tan x的值.(2)求 cos2x/(√2cos(π/4+x)
已知f(x) =sin^2 x +2sinx cos x + 3 cos^2 x ,x∈(0 ,π) .
三角函数题.已知tanx=2.求(sin(π-x)cos(2π-x)sin(-x+3π/2))/tan(-x-π)sin
求证 tan(2π-X)sin(-2π-X)cos(6π-X)/ sin(X+3π/2)*cos(X+3π/2)=-ta