Question

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

若cos(-a-π/2)=1/5,a∈[π,3π/2]，求f(a)的值

Answer 1

∵a∈[π,3π/2]
∴sina<0 cosa<0
cos(-a-π/2)=1/5
cos(a+π/2)=1/5
-sina=1/5
sina=-1/5
cosa=-√(1-sin^2a)
=-√[1-(-1/5)^2]
=-2√6/5
f(a)=sin(a-3π)cos(2π-a)sin(-a+3π/2)/[cos(-π-a)sin(-π-a)]
=-sin(π-a)cos(-a)[-cos(-a)]/{cos(π+a)[-sin(π+a)]}
=-sinacosa(-cosa)/[(-cosa)sina]
=-cosa
=-(-2√6/5)
=2√6/5