已知向量a (cosx,-根号3 2),向量b=sin,cos2x

(1)F(x)=2(cosx)^2-2√3cosxsinx=cos2x-√3sin2x+1=2sin(2x-π/6)+1最小正周期T=π函数f（x）在【0,π】上的单调递增区间为[0,π/3]∪[5π

f(x)=向量a乘向量b=2sinx*√3cosx+(√2cosx+1)(√2cosx-1)=√3sin2x+2(cosx)²-1=√3sin2x+cos2x=2sin(2x+π/6)∴T=

已知函数f(x)=向量a*向量b,其中向量a=（2cosx,根号3sinx),向量b=(cosx,-2cosx)（1）求函数f（x）在[0,π]上的单调递增区间和最小值（2）在三角形ABC中,a、b、

f(x)=2[√3cosxsinx+2(cosx)^2]-2[(sinx)^2+(2cosx)^2]-1=√3sin2x-2(cosx)^2-3=√3sin2x-cos2x-4=2sin(2x-π/6

f（x）=a.b=sinxcosx+√3cosx^2=1/2sin2x+√3(cos2x+1)/2=sin(2x+π/3)+√3/2(1)递增区间：-π/2+2kπ

向量a=(sinx,根号3cosx),向量b=(cosx,cosx),f(x)=向量a*向量b=sinxcosx+√3cos²x=1/2sin2x+√3/2(1+cos2x)=1/2sin2

f(x)=5根号3*sinxcosx+2(cosx)²+(sinx)²+4(cosx)²=5根号3sinxcosx+5(cosx)²+1=5*根号3/2*sin

题目写得应该稍有问题,我想应该是：函数f(x)=2*a点乘b+(2m-1)要不然f(x)也是个向量.1)f(x)=2(isqrt(3)sinx+jcosx).(icosx+jcosx)+2m-1=2s

f(x)=mn=2cos^2x+2√3sinxcosx+a-1+1=cos2x+√3sin2x+a+1=2sin(2x+π/6)+a+1f(x)=0sin(2x+π/6)=(-a-1)/2f(x)在【

1a·b=√3sinxcosx+cosx^2=√3sin(2x)/2+(1+cos(2x))/2=sin(2x+π/6)+1/2故：f(x)=a·b-1/2=sin(2x+π/6)最小正周期：T=2π

向量a.向量b=√3sinxcosx+cos^2x-m^2.=2[(√3/2)sin2x+(1+cos2x)/2-m^2.(1)f(x)=sin(2x+π/6)+1/2-m^2.再问：还有第二问呢？再

(1)向量a=（2cosx,根号3）,b=（cosx,-sinx）a∥b,所以%D¢cosx/cosx=√3/(-sinx)%D%A即%D%Asinx=-√3/2%D%A所以%D¢

f(x)=2sinxcosx+2√3(cosx)^2-1-√3=sin2x+√3cos2x-1=2sin(2x+π/3)-1(1)当2x+π/3=π/2,即x=π/12时,f(x)取得最大值f(π/1

f(x)=2cos²x+2√3sinxcosx=cos2x+1+√3sin2x=2sin(2x+π/6)+1(1)T=2π/2=π(2)当x∈[π/24,5π/24]时,2x+π/6∈[π/

(1)f(x)=cos²x+√3sinxcosx=(1+cos2x)/2+√3/2sin2x=√3/2sin2x+1/2cos2x+1/2=sin(2x+π/6)+(1/2)最小正周期为π（

1.f(x)=2(√3sinxcosx+(cosx)^2)+2m-1=√3sin2x+cos2x+2m=2sin(2x+pi/6)+2m最小正周期=pi2.x属于[0,pi/2]f(x)最小值=2si

（1）解析：函数f(x)=√3cos²x+sinxcosx=√3(cos2x+1)/2+1/2sin2x=sin(2x+π/6)+√3则函数的最小正周期为π,图像的对称轴方程x=kπ+π/6

紧跟题意1、f(x)=a*b=-cos^2x+sinx*根号3*cosx=sin(2x-π/6)-1/22、T=2π/2=π,-π/2+2kπ《2x-π/6《π/2+2kπ,解得kπ-π/6《x《π/

祝你开心!希望能帮到你,如果不懂,请追问,祝学习进步!O(∩_∩)O再问：���ʵ�һ�ʵĵڶ�������ô����再答：���ǹ�ʽ��sin2x=2sinxcosx��cos2x=1-2sin&

(1)向量a=（2cosx,根号3）,b=（cosx,-sinx）a∥b,所以2cosx/cosx=√3/(-sinx)即sinx=-√3/2所以2cos²x-sinx=2(1-sin