题目内容

求下列各代数式的值
(1)
1-2sin10°cos10°
sin170°-
1-sin2170°
          
(2)cos50°(
3
-tan10°
(1)原式=
1-2sin10°cos10°
sin170°-
1-sin2170°

=
(sin10°-cos10°)2
sin10°-
cos210°

=
cos10°-sin10°
sin10°-cos10°

=-1…(6分)
(2)原式=cos50°(
3
-tan10°

=cos50°(
3
-
sin10°
cos10°
)

=cos50°×
3
cos10°-sin10°
cos10°

=cos50°×
sin(60°-10°)
cos10°

=
sin100°
cos10°

=-1…(12分)
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网