题目内容

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

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