已知x为第三象限角,且{ f(x)=sin(π-x)cos(2π-x)tan(-x+3π/2) }/{cotxsin(π+x)}
问题描述：
已知x为第三象限角,且{ f(x)=sin(π-x)cos(2π-x)tan(-x+3π/2) }/{cotxsin(π+x)}
1.化简f(x)
2.若cos(x-3π/2)=1/5,求f(x)的值.
3.若x=-1860°,求f(x)的值.
答
f(x)=sin(π-x)cos(2π-x)tan(-x+3π/2) }/{cotxsin(π+x)=f(x)=sinxcosxcotx }/-{cotxsinx}=cos^2x/-cosx=-cosxcos(x-3π/2)=1/5=-sinxsinx=-1/5cosx=2√6／5f(x)=-2√6／5x=-1860°f(x)=-cosx=-cos（-1860°）=-co...