问题 解答题
求下列各代数式的值
(1)
1-2sin10°cos10°
sin170°-
1-sin2170°
          
(2)cos50°(
3
-tan10°
答案

(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)

单项选择题 A3/A4型题
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