已知为x第三象限角,f(x)=sin(x-π/2)cos(3π+x)tan(π-x)/tan(-x-π)sin(-x-π
已知x为第三象限角,且{ f(x)=sin(π-x)cos(2π-x)tan(-x+3π/2) }/{cotxsin(π
已知a为第三象限角,且f(a)=sin(π-x)cos(2π-x)tan(-x+π)/sin(π+x)
已知f(x)=(sin(π-x)*cos(2π-x))/cos(-π-x)*tan(π-x),则f(-31π/3)的值为
已知(x)=[sin(π-x)cos(2π-x)tan(-x+π)]/[cos(-π/2+x)]
已知f(x)=sin(π-x)cos(2π-x)tan(-x+π)/cos(-π/2+x),求f(-31π/3)的值
已知tan=2,求(cos x+sin x)/(cos x-sin x)+sin^2x
f(a)=[sin(π-x)cos(2π-x)tan(-x+3π/2)]/cot(-x -π)sin(-π--x) 化简
判断下列函数的奇偶性 f(x)=[sin(π/2+x)cos(π/2-x)tan(-x+3π)]/[sin(7π-x)t
三角函数已知f(x)=(2cos^4x—2cos²x+1/2)/(2tan(π/4—x)sin²(x
三角函数题.已知tanx=2.求(sin(π-x)cos(2π-x)sin(-x+3π/2))/tan(-x-π)sin
已知sin(π/2-x)+sin(π-x)/cos(-x)+sin(2π-x)=2009,则tan(x+5π/4)等于?
f(x)=1/cos²x-tan²x+根号2*sin(2x-π/4)