已知a为第三象限角,且f(a)=sin(π-x)cos(2π-x)tan(-x+π)/sin(π+x)

题目

已知a为第三象限角,且f(a)=sin(π-x)cos(2π-x)tan(-x+π)/sin(π+x)
1.化简f(a)
2.若cos(a-3π/2)=1/5,求f(a)

答案

(1)
f(x)=sin(π-x)cos(2π-x)tan(-x+π)/sin(π+x)
=(sinx)(cosx)(-tanx)/(-sinx)
= cosxtanx
=sinx
(2)
cos(x-3π/2)=1/5
-sinx =1/5
f(x) = sinx = -1/5