已知COSx(a π 4)=-根号10 10,a∈(0,π 2

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=

(1)|a|=√(sin²＋(-cosx)²)=1|b|=√(cos²x＋(√3cosx)²)=2cosx=1再问：第二问主要再答：再答：S△ABC=1/2ab

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

见图片

(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）f（x）＝ab=2√2cosx+sinxcosx+2√2sinx-sinxcosx=2√2(sinx+cosx)=2√2×√2sin(x+π/4)=4sin(x+π/4)x属于【0,π】,(x+

（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

已知,a=(-cosx,sinx),b=(cosx,√3cosx),所以函数f(x)=a·b=-cos²x+√3sinxcosx=-cos2x/2+√3sin2x/2-1/2=sin(2x-

f(x)=cos²x+√3sinxcosx=1/2(2cos²x-1)+√3sinxcosx+1/2=1/2cos2x+√3/2sin2x+1/2=cos(2x-π/3)+1/2当

(1)a*b=0sin2x-cos2x=0sqr(2)sin(2x-π/4)=0x=π/8+kπ/2,k∈Z(2)f(x)=sqr(2)sin(2x-π/4)x∈(3π/8+kπ,7π/8+kπ),k

∵a*b=8/5∴√2cosx+√2sinx=8/5sinx+cosx=4√2/5两边平方得,2sinxcosx=7/25∵π/4再问：但是答案是-28/75再答：呵呵，算错了！sinx-cosx=√

f(x)=sin(2x-π/3)-√3/2所以函数的最小正周期为π对称中心的坐标为(kπ/2,-√3/2)值域为[-√3,1-√3/2]再问：求值sin(60°-α/2)×cos(30°-α/2)×(

1）x=π/6,则a=(√3/2,1/2)b=(-√3/2,3/2),令c=na+mb,解得n=-√3/2,m=√3/6,所以c=(-√3/2)a+(√3/6)b2）f(x)=2(cosx)^2+√3

f(x)=a*b=sin(π/2+x）*sinx+√3cosx*cosx=1/2sin2x+√3/2（1+cos2x）=sin（2x+π/3）+√3/2最小正周期T=2π/2=π单调递增区间：2x+π

(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

(1)f(x)=sin(2x+π/6)+√3最小正周期为π,对称轴方程x=kπ+π/6,x=kπ-π/3(2)解析：f(-5π/12)=-sin(4π/6)+√3=√3/2,f(π/12)=sin(π

f(x)=√3sinxcosx+(cosx)^2+m=sin(2x+π/6)+m+1/21.最小正周期T=2π/2=π2.x∈[-π/6,π/3]2x+π/6∈[-π/6,5π/6]可见x=-π/6时