问题 解答题
已知6sin2α+sinαcosα-2cos2α=0,α∈[
π
2
,π]
,求sin(2α+
π
3
)
的值.
答案

由已知得:(3sinα+2cosα)(2sinα-cosα)=0⇔3sinα+2cosα=0或2sinα-cosα=0

由已知条件可知cosα≠0,所以α≠

π
2
,即α∈(
π
2
,π)
.于是tanα<0,∴tanα=-
2
3

sin(2α+

π
3
)=sin2αcos
π
3
+cos2αsin
π
3

=sinαcosα+

3
2
(cos2α-sin2α)

=

sinαcosα
cos2α+sin2α
+
3
2
×
cos2α-sin2α
cos2α+sin2α

=

tanα
1+tan2α
+
3
2
×
1-tan2α
1+tan2α
.

将tanα=-

2
3
代入上式得

sin(2α+

π
3
)=-
(-
2
3
)
1+(-
2
3
)
2
+
3
2
×
1-(-
2
3
)
2
1+(-
2
3
)
2

=-

6
13
+
5
26
3
.即为所求.

阅读理解

阅读理解。

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