Question

已知函数f(x)=cosx(根号3sinx+cosx)-1/2(x∈R)。

（1）求函数f(x)的最小正周期及在区间【0，π/2】上的最大值和最小值。（2）若f(x0)=5/13,x0∈【π/4，π/2】，求cos2x0的值。

Answer 1

1、
f(x)=(√3/2)sin2x+(1/2)cos2x
=sin(2x+π/6)
周期T=2π/2=π
当x∈【盯祥0，π/2】时，
2x+π/6∈【π/6，7π/6】
则sin(2x+π/6)∈【-1/2，1】
所以，f(x)在区间【0，π/2】上的最大值为1，最小值伍前为-1/2；

2、
即：sin(2x0+π/6)=5/13
x0∈【π/4，π/2】，则2x0+π/6∈【2π/3，7π/6】
所以，cos(2x0+π/6)=-12/13
cos2x0=cos[(2x0+π/6)-π/6]
=cos(2x0+π/凯橘搏6)cos(π/6)+sin(2x0+π/6)sin(π/6)
=(-12/13)(√3/2)+(5/13)(1/2)
=(5-12√3)/26

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Answer 2

解：
(1)
f(x)=cosx(√3sinx+cosx) -1/2
=√3sinxcosx +cos²x -1/2
=(√3/2)sin(2x)+[1+cos(2x)]/液悄2 -1/2
=(√3/2)sin(2x)+ (1/闹扒2)cos(2x)
=sin(2x+π/6)
Tmin=2π/2=π
x∈[0，π/2]
π/6≤2x+π/6≤7π/6
-1/2≤sin(2x+π/6)≤1
f(x)max=1 f(x)min=-1/2
(2)
x0∈[π/4，π/2]
2π/3≤2x0+π/6≤7π/6
f(x0)=sin(2x0+π/6)=5/13>0
2π/3≤2x0+π/6<π
cos(2x0+π/6)<0
cos(2x0+π/6)=-√[1-sin²(2x0+π/6)]=-√[1-(5/13)²]=-12/13
cos(2x0)=cos[(2x0+π/6)-π/6]
=cos(2x0+π/6)cos(π/6)+sin(2x0+π/6)sin(π/6)
=(-12/13)(√液埋昌3/2) +(5/13)(1/2)
=(5-12√3)/26

Answer 3

f(x)=cosx(√3sinx+cosx)-1/2=√3/2sin2x+(cos2x+1)/2-1/2=sin(2x+π/6)
最小正周期T=π,最大值y=1,最小值y=-1
（2）若f(x0)=5/13,x0∈【戚闭π/4，π/2】，求祥岁cos2x0的值。
sin(2x0+π/6)=5/13=√3/2sin2x0+1/2cos2x0 2x0=u
5/13=√3/2sinu+1/2cosu , 5/13-1/2cosu=√3/2sinu
25/169-5/13cosu+1/4cos^2u=3/4-3/4cos^2u
cos^2u-5/13cosu-407/676=0, (cosu-5/26)^2-108/169=0, cosu=5/26±6√3/13
x0∈【π/谨仔睁4，π/2】，2x0∈【π/2，π】，cos2x0=(5-12√3)/26