1.f(x)=(sin2x+cos2x)/(tanx+cotx)求：最小正周期 和值域.2.f(x)=2sin x/4 cos x/4 - 2根号3sin^2x/4 +根号3求：①T和最值.② 令g(x)=f(x+π/3),判断g(x)的奇偶性,说明理由.3.已知 f(x)=cos^2(x+π/12),g(x）=1+ 0.5sin2x①设x=x0

1.f(x)=(sin2x+cos2x)/(tanx+cotx)求：最小正周期 和值域.2.f(x)=2sin x/4 cos x/4 - 2根号3sin^2x/4 +根号3求：①T和最值.② 令g(x)=f(x+π/3),判断g(x)的奇偶性,说明理由.3.已知 f(x)=cos^2(x+π/12),g(x）=1+ 0.5sin2x①设x=x0
1.f(x)=(sin2x+cos2x)/(tanx+cotx)
求：最小正周期 和值域.
2.f(x)=2sin x/4 cos x/4 - 2根号3sin^2x/4 +根号3
求：①T和最值.
② 令g(x)=f(x+π/3),判断g(x)的奇偶性,说明理由.
3.已知 f(x)=cos^2(x+π/12),g(x）=1+ 0.5sin2x
①设x=x0 是函数y=f(x)图像的一条对称轴,求g(x0)
②求函数h(x)=f(x)+g(x)的单调递增区间.

1.f(x)=(sin2x+cos2x)/(tanx+cotx)求：最小正周期 和值域.2.f(x)=2sin x/4 cos x/4 - 2根号3sin^2x/4 +根号3求：①T和最值.② 令g(x)=f(x+π/3),判断g(x)的奇偶性,说明理由.3.已知 f(x)=cos^2(x+π/12),g(x）=1+ 0.5sin2x①设x=x0
1.f(x)=(sin2x+cos2x)/(tanx+cotx)
=(sin2x+cos2x)/(sinx/cosx+cosx/sinx)
=(sin2x+cos2x)/[(si'n^x+cos^x)/sinxcosx]
=(sin2x+cos2x)/(1/sinxcosx)
=(sin2x+cos2x)sinxcosx
=(1/2)(sin2x+cos2x)sin2x
=(1/4)[2sin^(2x)+2cos2xsin2x]
=(1/4)[1-cos4x+sin4x]
=1/4-根号2*sin(4x-∏/4]/4 ,x≠k∏/2, 4x≠2k∏
-1<=sin(4x-∏/4]<=1
根号2/4<=-根号2*sin(4x-∏/4]/4<=根号2/4,
所以(1-根号2)/4<=1/4-根号2*sin(4x-∏/4]/4<=(1+根号2)/4
所以f(x)的值域[(1-根号2)/4,(1+根号2)/4]
周期2∏/4=∏/2
2.
f(x)=sin(x/2)+√3cos(x/2)
=2sin(x/2+arctan√3)
=2sin(x/2+π/3)
T=2π/(1/2)=4π
g(x)=f(x+π/3)=2sin[(x+π/3)/2+π/3]
=2sin(x/2+π/2)
=2cos(x/2)
g(-x)=2cos(-x/2)=2cos(x/2)=g(x)
且定义域x属于R,关于原点对称
所以是偶函数
3.
(1)∵cos2x=2cos^2x-1
∴f(x)=1/2+cos(2x+π/6)/2
对称轴2x0+π/6=π+2kπ
x0=5π/12+kπ
g(x0)=1+1/2sin(5π/6+2kπ)=5/4
（2）h(x)=1/2+cos(2x+π/6)/2+1+1/2sin2x
=3/2+0.5*(√3/2*cos2x-1/2sin2x+sin2x)
=3/2+0.5*sin(2x+π/3)
递减区间（π/12+kπ,7π/12+kπ）

1.已知函数f(x)=(sin2x+cos2x)/(tanx+cotx),求f(x)的值域.和周期
f(x)=(sin2x+cos2x)/(tanx+cotx)
=(sin2x+cos2x)/(sinx/cosx+cosx/sinx)
=(sin2x+cos2x)/[(si'n^x+cos^x)/sinxcosx]
=(sin2x+cos2x)/(1/sinx...

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1.已知函数f(x)=(sin2x+cos2x)/(tanx+cotx),求f(x)的值域.和周期
f(x)=(sin2x+cos2x)/(tanx+cotx)
=(sin2x+cos2x)/(sinx/cosx+cosx/sinx)
=(sin2x+cos2x)/[(si'n^x+cos^x)/sinxcosx]
=(sin2x+cos2x)/(1/sinxcosx)
=(sin2x+cos2x)sinxcosx
=(1/2)(sin2x+cos2x)sin2x
=(1/4)[2sin^(2x)+2cos2xsin2x]
=(1/4)[1-cos4x+sin4x]
=1/4-根号2*sin(4x-∏/4]/4 ,x≠k∏/2, 4x≠2k∏
-1<=sin(4x-∏/4]<=1
根号2/4<=-根号2*sin(4x-∏/4]/4<=根号2/4,
所以(1-根号2)/4<=1/4-根号2*sin(4x-∏/4]/4<=(1+根号2)/4
所以f(x)的值域[(1-根号2)/4,(1+根号2)/4]
周期2∏/4=∏/2
2.
f(x)=2sin x/4 cos x/4 - 2根号3sin^2x/4 +根号3
f(x)=sin x/2+√3(cos x/2+1)+√3
f(x)=sin x/2+√3cos x/2
f(x)=2sin（x/2+∏/3)

由求周期的公式
T=2π/2=π
正弦函数的最值分别是-1和1
所以-1≤sin(2x+π/3)≤1
所以-2≤f(x)≤2
最大值=2，最小值=-1
g(x)=f(x+π/3)=2sin[(x+π/3)/2+π/3]
=2sin(x/2+π/2)
=2cos(x/2)
g(-x)=2cos(-x/2)=2cos(x/2)=g(x)
且定义域x属于R，关于原点对称
所以是偶函数
3.已知 f(x)=cos^2(x+π/12),g(x）=1+ 0.5sin2x
①设x=x0 是函数y=f(x)图像的一条对称轴，求g(x0)
②求函数h(x)=f(x)+g(x)的单调递增区间。
1)对称轴应该不难吧,先化简,原式=cos^2 (X+兀/12)-0.5+0.5=.5cos(2X+兀/6)-.5,对称轴在2X+兀/6=K兀,即X。=K兀/2-兀/12,代入GX,1+.5sin(K兀-兀/6)=0.5或1.5.
2)FX+GX=0.5+.5cos(2X+兀/6)+.5sin2X,把cos(2X+兀/6)化出来=根3/2cos2X-1/2sin2X 即原式=0.5+根3/4cos2X+1/4sin2X=0.5+1/4(2sin(2X+兀/3),当2X+兀/3在[-兀/2,兀/2]+2K兀,即...(这个自己解吧,我手头没计算器)上递增.

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