问题 解答题
(1)化简
sin(-α)cos(2π+α)
sin(
π
2
+α)

(2)计算4
1
2
+2log23-log2
9
8

(3)已知tanθ=3,求
1
sin2θ-2sinθcosθ
的值.
答案

(1)原式=

-sinαcosα
cosα
=-sinα;

(2)原式=2+log232-log2

9
8
=2+log223=2+3=5;

(3)∵tanθ=3,∴原式=

sin2θ+cos2θ
sin2θ-2sinθcosθ
=
tan2θ+1
tan2θ-2tanθ
=
9+1
9-2
=
10
7

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