已知函数

.
(1)求函数

的最小正周期及单调递增区间;
(2)需要把函数

的图像经过怎样的变换才能得到函数

的图像?
(3)在

中,

、

、

分别为三边
、
、
所对的角,若

,

,求

的最大值.
解:(1)

(2分)
最小正周期为

,由

(
k
Z)可得

(
k
Z)
即函数的单调递增区间为

(
k
Z)
(2)要得到函数

的图像只需把函数

的图像经过以下变换得到:①把函数

横坐标伸长为原来的2倍,纵坐标不变得到函数

的图像;②再把函数

的图像纵坐标缩短为原来的

,横坐标不变,得到函数

的图像;③再把函数

的图像向左平移

个单位得到

的图像
(3)由

可得

,即

,又0<

<

,所以

.由余弦定理可得

,即

(11分),即

.又

,所以

故

故当且仅当

,即

时,

取得最大值

.
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