问题 解答题
(1)求值:
1-2sin190°cos170°
cos170°+
1-cos2190

(2)已知sinα+cosα=
4
5
π
2
<α<π,求sinα-cosα.
答案

(1)

1-2sin190°cos170°
cos170°+
1-cos2190

=

1-2sin10°cos10°
-cos10°+
1-cos210

=

(sin10°-cos10°)2
sin10°-cos10°

=

cos10°-sin10°
sin10°-cos10°

=-1.

(2)∵sinα+cosα=

4
5

∴(sinα+cosα)2=

16
25

2sinαcosα=-

9
25

∴(sinα-cosα)2

=1-2sinαcosα

=

34
25

π
2
<α<π,

∴sinα-cosα=

34
5

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