> 数学 >
已知函数f(x)=sin2(x-
π
6
)+cos2(x-
π
3
)+sinx•cosx,x∈R.
(1)求f(x)的最大值及取得最大值时的x的值;
(2)求f(x)在[0,π]上的单调增区间.
人气:408 ℃ 时间:2019-10-04 11:19:35
解答
(1)由题意得,
f(x)=(sinxcos
π
6
−cosxsin
π
6
)2
+(cosxcos
π
3
+sinxsin
π
3
)
2
+sinx•cosx
=sin2x+sinx•cosx+
1
2
=
1
2
(sin2x−cos2x)
+1
=
2
2
sin(2x−
π
4
)+1

2x−
π
4
=
π
2
+2kπ
(k∈Z),
即x=
8
+kπ
(k∈Z)时,函数f(x)取最大值为:
2
2
+1

(2)由0≤x≤π得,-
π
4
≤2x−
π
4
4

∴函数f(x)=
2
2
sin(2x−
π
4
)+1
的增区间是:[-
π
4
π
2
].
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