Question

已知f(a)=sin(a-3π)cos(2π-a)sin(-a+3π/2)/cos（-π-a)sin(-π-a)

1.若a是第三象限的角，且cos（a-3π/2）=1/5，求f(a)的值
2.若a=-31π/3，求f(a)的值
在线等 谢谢

Answer 1

f(a)=sin(a-3π)cos(2π-a)sin(-a+3π/2)/cos（-π-a)sin(-π-a)
=(-sina)cosa(-cosa)/(-cosa)sina
=-cosa
1 cos(a-3π/2)=1/5
-sina=1/5
sina=-1/5
cosa=±2根号下5/5
a为第三象限角 所以cosa<0 cosa=-2根号下5/5
f(a)=-cosa=2根号下5/5
2 f(-31π/3)=-cos(-31π/3)
=-cos(-10π-π/3)
=-cosπ/3
=-1/2

Answer 2

(1)
cos（a-3π/2）=1/5
cos（3π/2-a）=1/5
sina=-1/5
cosa = -2√6/5
f(a)
= sin(a-3π)cos(2π-a)sin(-a+3π/2)/[cos(-π-a)sin(-π-a)]
= (-sina)(cosa)(-cosa)/ [(-cosa)(sina)]
=-cosa
=2√6/5
(2)
a=-31π/3
f(a) = -cos(-31π/3)
=-cos(31π/3)
=-cos(10π+π/3)
= -cosπ/3
=-1/2