高中数学已知函数f(x )=cos^2 x-√3 sin x cosx+2sin^2x-1/21-求函数fx最小正周期2-若x属于[0,pai/2],求函数fx的值域
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高中数学已知函数f(x )=cos^2 x-√3 sin x cosx+2sin^2x-1/21-求函数fx最小正周期2-若x属于[0,pai/2],求函数fx的值域
高中数学已知函数f(x )=cos^2 x-√3 sin x cosx+2sin^2x-1/2
1-求函数fx最小正周期
2-若x属于[0,pai/2],求函数fx的值域
高中数学已知函数f(x )=cos^2 x-√3 sin x cosx+2sin^2x-1/21-求函数fx最小正周期2-若x属于[0,pai/2],求函数fx的值域
答:
f(x)=(cosx)^2-√3sinxcosx+2(sinx)^2-1/2
f(x)=(1/2)*cos(2x) -(√3/2)sin2x +1-cos(2x)
f(x)=-(√3/2)sin2x-(1/2)cos(2x)+1
f(x)= - sin(2x+π/6) +1
1)
f(x)的最小正周期T=2π/w=2π/2=π
2)
0<=x<=π/2
0<=2x<=π
π/6<=2x+π/6<=7π/6
所以:-1/2<=sin(2x+π/6)<=1
所以:1-1/2<=f(x)<=1+1
所以:1/2<=f(x)<=2
所以:f(x)的值域为[ 1/2,2]
1
f(x)=(1/2)[2cos^2(x)-1]-√3/2sin2x
=(1/2)cos2x-(√3/2)sin2x
=(cos2x)(cosπ/3)-sin2xsin(π/3)
=cos(2x+π/3)
由周期公式得:
T=2π/2=π
0≤x≤π/2
π/3≤2x+π/3≤4π/3
函数y=...
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1
f(x)=(1/2)[2cos^2(x)-1]-√3/2sin2x
=(1/2)cos2x-(√3/2)sin2x
=(cos2x)(cosπ/3)-sin2xsin(π/3)
=cos(2x+π/3)
由周期公式得:
T=2π/2=π
0≤x≤π/2
π/3≤2x+π/3≤4π/3
函数y=cost在[π/3,4π/3]上的单调性是;先减后增,且跃过最小值;右端点为负值;
f(max)=f(0)=1/2
f(min)=-1
原函数的值域为:[-1,1/2]
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f(x )=cos^2 x-√3 sin x cosx+2sin^2x-1/2
=1/2+1/2cos2x-√3 /2sin 2x+(1-cos2x)-1/2
=1-(1/2cos2x+√3 /2sin 2x)
=1-cos(2x-π/3)
=1+sin[π/2+(2x-π/3)]
=1+sin(2x+π/6)
T=2π/2=π
x属于[0,pai/2]