问题 解答题
求值
(1)sin2840°+cos540°+tan225°-cos(-330°)+sin(-210°)
(2)已知tanβ=
1
2
,求sin2β-3sinβcosβ+4cos2β的值.
答案

(1)∵sin2840°+cos540°+tan225°-cos(-330°)+sin(-210°)

=sin2120°+cos180°+tan45°-cos30°+sin150°

=

3
4
-1+1-
3
2
+
1
2

=

5-2
3
4

(2)∵tanβ=

1
2

∴sin2β-3sinβcosβ+4cos2β

=

sin2β-3sinβcosβ+4cos2β
sin2β+cos2β

=

tan2β-3tanβ+4
tan2β+1

=

11
5

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