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+
)3π 2 cos(−π−α)sin(−π−α)
(1)化简f(α);
(2)若α=-
,求f(α)的值. 31π 3
答
(1)f(α)=
sin(α−3π)cos(2π−α)sin(−α+
)3π 2 cos(−π−α)sin(−π−α)
=
−sin(3π−α)•cos(−α)•(−cosα) cos(π+α)•[−sin(π+α)]
=
=-cosα;−sinα•cosα•(−cosα) −cosα•sinα
(2)∵α=-
,f(α)=-cosα,31π 3
∴f(α)=-cos(−
)31π 3
=-cos
=-cos(10π+31π 3
)=-cosπ 3
=-π 3
.1 2