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。...
(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。
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1、
f(x)=(√3/2)sin2x+(1/2)cos2x
=sin(2x+π/6)
周期T=2π/2=π
当x∈【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/6)cos(π/6)+sin(2x0+π/6)sin(π/6)
=(-12/13)(√3/2)+(5/13)(1/2)
=(5-12√3)/26
祝你开心!希望能帮到你,如果不懂,请追问,祝学习进步!O(∩_∩)O
f(x)=(√3/2)sin2x+(1/2)cos2x
=sin(2x+π/6)
周期T=2π/2=π
当x∈【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/6)cos(π/6)+sin(2x0+π/6)sin(π/6)
=(-12/13)(√3/2)+(5/13)(1/2)
=(5-12√3)/26
祝你开心!希望能帮到你,如果不懂,请追问,祝学习进步!O(∩_∩)O
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解:
(1)
f(x)=cosx(√3sinx+cosx) -1/2
=√3sinxcosx +cos²x -1/2
=(√3/2)sin(2x)+[1+cos(2x)]/2 -1/2
=(√3/2)sin(2x)+ (1/2)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)(√3/2) +(5/13)(1/2)
=(5-12√3)/26
(1)
f(x)=cosx(√3sinx+cosx) -1/2
=√3sinxcosx +cos²x -1/2
=(√3/2)sin(2x)+[1+cos(2x)]/2 -1/2
=(√3/2)sin(2x)+ (1/2)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)(√3/2) +(5/13)(1/2)
=(5-12√3)/26
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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∈【π/4,π/2】,求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/4,π/2】,2x0∈【π/2,π】,cos2x0=(5-12√3)/26
最小正周期T=π,最大值y=1,最小值y=-1
(2)若f(x0)=5/13,x0∈【π/4,π/2】,求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/4,π/2】,2x0∈【π/2,π】,cos2x0=(5-12√3)/26
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