已知f(α)=sin(α−3π)cos(2π−α)sin(−α+3π2)cos(−π−α)sin(−π−α). (1)化简f(α); (2)若α=-31π3,求f(α)的值.

问题描述:

已知f(α)=

sin(α−3π)cos(2π−α)sin(−α+
2
)
cos(−π−α)sin(−π−α)

(1)化简f(α);
(2)若α=-
31π
3
,求f(α)的值.

(1)f(α)=

sin(α−3π)cos(2π−α)sin(−α+
2
)
cos(−π−α)sin(−π−α)

=
−sin(3π−α)•cos(−α)•(−cosα)
cos(π+α)•[−sin(π+α)]

=
−sinα•cosα•(−cosα)
−cosα•sinα
=-cosα;
(2)∵α=-
31π
3
,f(α)=-cosα,
∴f(α)=-cos(
31π
3

=-cos
31π
3
=-cos(10π+
π
3
)=-cos
π
3
=-
1
2