Question

求下列三角函数式的值.

(1)sin495°·cos(-675°);

(2)3sin(-1 200°)·tan(-)-cos585°·tan().

Answer 1

解析:(1)sin495°·cos(-675°)

=sin(135°+360°)·cos675°

=sin135°·cos315°

=sin(180°-45°)·cos(360°-45°)

=sin45°·cos45°

=×=.

(2)sin(-1 200°)·tan(-)-cos585°·tan()

=-sin1 200°·(-)-cos(720°-135°)·tan(-8π-)

=sin(1 080°+120°)-cos135°·tan(-)

=-(-)·(-1)

=.

答案:(1) ;(2) .