Question

已知函数f（x）=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx

Answer 1

(1).
f(x)=2cosx*sin(x+π/3)-√李知3sin^2x+sinx*cosx=2cosx(1/2sinx+√3/2cosx)-√3sin^2x+sinxcosx
=2sinxcosx+√3cos^2x-√3sin^2x=sin2x+√3cos2x=2(1/2sin2x+√3/猛册2cos2x)=2sin(2x+π/3)
令π/2+2kπ(2).
f(x)平移后得到：g(x)=f(x-m)=2sin(2x-2m+π/3)，求哪知消解后面的问题。（后面的问题没说得很清楚，没有所谓最小正值，最大值是2）。
觉得好，请采纳，谢谢。

Answer 2

1.
f（x）=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx
=
2cosx*sin(x+π/3)-
2sinx*[(√3/2)sinx-(1/2)cosx]
=
2cosx*sin(x+π/3)-
2sinx*[sin(π/3)sinx-cos(π/3)cosx)]
=
2cosx*sin(x+π/3)+
2sinxcos(x+π/3)
=
2sin(2x+π/3)
f(x)的单调区间
2kπ
-
π/蔽伍冲2
<=
2x+π/3
<
2kπ
+
π/2
单调增
(2k+1)π
-
π/2
<=
2x+π/3
<
(2k+1)π
+
π/2
单调减
k=
0
,
+/-1,
+/-
2,
+/-
3
......
=>
[2kπ
-
π/2
-
π/3]/2
<=
x
<
[2kπ
+
π/2
-
π/3]/2
[(2k+1)π
-
π/2
-π/3]/2
<=
x
<
[(2k+1)π
+
π/2
-
π/3]/2
2.
f(x)
平移
a(m,0)
即沿x
轴橘笑向正方向移动
g(x)=f(x-m)
=
2sin[2(x-m)+π/3]
g(x)
为
偶函数宏歼
即
g(x)=
g(-x)
=>
2sin[2(x-m)+π/3]
=
2sin[2(-x-m)+π/3]
又sin为奇函数
=>
2sin[2(x-m)+π/3]
=
-2sin{-[2(-x-m)+π/3]}
=>
2sin[2(x-m)+π/3]
=
2sin{[2(x+m)-π/3]}
=>
2(x-m)
+
π/3
+
2kπ
=
2(x+m)-π/3
=>
4m=
2π/3
+
2kπ
m=π/6
取最小正值(k=0)