已知函数f(x)=sin^2x+2sinxcosx-cos^2x,x∈R
(1)函数f(x)的最小正周期
(2)函数f(x)取最大值时x的取值集合
(3)函数f(x)的图像可由函数y=sinx,x∈R的图像经过怎样的变换得到
已知函数f(x)=sin^2x+2sinxcosx-cos^2x,x∈R
你好:(1)。f(x)=0.5(1-cos2x)+sin2x-0.5(1+cos2x)=sin2x-cos2x=√2sin(2x-π\/4)所以最小正周期为:T=2π\/2=π (2).因为函数y=sinx在x=2kπ+π\/2处取最大值,所以:令2x-π\/4=2kπ+π\/2 x=kπ+3π\/4 k为整数 (3).可以由y=sinx,x∈R先把横坐标变为原...
已知函数f(x)=sin^2x+2sinxcosx-cos^2x,x属于R
1.f(x)=(sinx)^2+2sinxcosx-(cosx)^2 =sin2x-cos2x =√2*sin(2x-π\/4)函数的最小正周期是T=2π\/2=π 2.函数的最大值是√2 当2x-π\/4=2kπ+π\/2(k∈Z)即x=kπ+3π\/8(k∈Z)时取得最大值 3.令2kπ-π\/2<2x-π\/4<2kπ+π\/2(k∈Z)得kπ-π\/8<x<kπ+3...
已知函数f(x)=sin^2x+2sinxcosx+3cos^2x,x属于R
=(2cos^2x-1)+2sinxcosx+2 =cos2x+sin2x+2 =v2sin(2x+π\/4)+2,——》f(x)最大=v2+2,此时,sin(2x+π\/4)=1——》2x+π\/4=2kπ+π\/2——》x=kπ+π\/8,k∈Z,正弦函数的周期是2kπ,即(2x+π\/4)整体的周期是2kπ,所以,针对自变量x的最小正周期为π,在计算正...
已知f(x)=sin²x+2sinxcosx+3cos²x,x∈R,求:函数f(x)的对称轴...
解f(x)=sin^2x+2sinxcosx+3cos^2x,=sin^2x+cos^2x+2sinxcosx+2cos^2x =1+sin2x+2cos^2x =2+sin2x+2cos^2x-1 =sin2x+cos2x+2 =√2sin(2x+π\/4)+2 函数的对称轴满足2x+π\/4=kπ+π\/2,k属于Z.故函数的对称轴为x=kπ\/2+π\/8,k属于Z.函数图像的对称中心满足2x+π\/4...
已知f(x)=sin^2x+2sinxcosx+3cos^2x x属于(0,π)求
f(x)=sin^2x+2sinxcosx+3cos^2x=(1—cos2x)\/2+sin2x+3\/2(1+cos2x)=cos2x+sin2x+2 =根号2sin(2x+π\/4)+2 (1)x属于(0,π),2x+π\/4属于(π\/4,9π\/4), f(x)最大值是根号2+2,2x+π\/4=π\/2时,即x=π\/8时,此函数有最大值 (2)π\/4<2x+π\/4≤π\/2或...
已知函数f(x)=sin^2x+2sinxcosx+cos^2x
f(x)=sin^2x+2sinxcosx+cos^2x=sin2x+1,最小值周期2π\/2=π,-1≤sin2x≤1,f(x)最大值2,最小值,0,[kπ-π\/4,kπ+π\/4]是递增[kπ+π\/4,kπ+3\/4π]递减
已知函数f(x)=sin^2x+2sinxcosx-3cos^2x 求函数的最小正周期和最大值...
解f(x)=sin^2x+2sinxcosx-3cos^2x =sin^2x+cos^2x +2sinxcosx-4cos^2x =1+2sinxcosx-4cos^2x =1+sin2x-4cos^2x =1+sin2x-2*(2cos^2x-1)-2 = 1+sin2x-2*(cos2x)-2 =sin2x-2cos2x-1 =√5(1\/√5sin2x-2\/√5cos2x)-1 =√5sin(2x-θ)-1 即函数的最小正周...
已知函数f(x)=sin^2x+2sinxcosx+3cos^2x
f(x)=sin^2x+2sinxcosx+3cos^2x =1+sin2x+2cos²x =2+sin2x+cos2x =2+√2sin(2x+π\/4)所以f(x)周期为Kπ (2)先纵向拉长√2个单位,得到f(x)=√2sinx 接着向上平移2个单位,得到f(x)=√2sinx+2 接着横坐标横向缩短一半,得到f(x)=2+√2sin(=2x 最后向左...
已知函数f(x)=2sinxcosx+2cos^2 x(x属于R),求f(x)的最小正周期;最小...
f(x)=sin2x+cos2x-1=√2sin(2x+π\/4)-1 f(x)的最小正周期是π 当x=2Kπ+3\/8π,K∈Z时,f(x)有最小值-1-√2
已知函数f(x)=sin²x+2sinxcosx+3cos²x,x∈R。求:⑴函数f(x)的...
f(x)=2sinxcosx+2(cosx)^2+1 =sin2x+cos2x+2 =√2sin(2x+45°)+2 ∴当 2x+45°=90°+360°k,k∈Z 即 x=22.5°+180°k时,f(x)取最大。当 -90°+360°k≤2x+45°≤90°+360°k,f(x)递增 得到 x∈[-67.5°+180°k,22.5°+180°k]...