已知向量a=(cosx,sinx),b=(根号2,根号2),若a*b=8 5且

已知向量a=（cosα,sinα）,向量b（cosx,sinx）,向量c=（sinx+2sinα,cosx+2cosα),其中0＜α＜x＜π；（1）若向量a与向量b的夹角为π/3,且向量a⊥向量c,求

（1）f(x)=2cos(2x+π/4),2x+π/4=π/2+kπ,k∈Z,解得x=π/8+k*π/2即（π/8+k*π/2,0）k∈Z（2）图略再问：请给出化简合并的过程。再答：cos^4*x-s

(1)α=45°f(x)=2sin(x+π/4)+sin2x因为0

(1)∵a=(2sin(π/4-x),cosx)b=(cos(π/4-x),2√3sinx)∴f(x)=2sin(π/4-x)*cos(π/4-x)+cosx*2√3sinx=sin[2(π/4-x)

f(x)=a*b+1=2cosxcosx+2cosxsin(π-x)+1=2cos^2x+2sinxcosx+1=cos2x+1+sin2x+1=cos2x+sin2x+2=√2sin(2x+π/4)

f(x)=2(cosx)^2+2根号3sinxcosx=cos2x+1+根号3sin2x=2sin(2x+Pai/6)+1单调增区间是:-Pai/2+2kPai

1)a-b=(-2cosx,2sinx/2-2cosx/2)f(x)=2+sinx-(1/4)[4cosx+4(sinx/2+cosx/2-2sinx/2cosx/2)]=2+sinx-cosx-(1

(1)∵b＝(cosx,sinx),c＝(sinx＋2sinα,cosx＋2cosα),α＝π4,∴f(x)＝b·c＝cosxsinx＋2cosxsinα＋sinxcosx＋2sinxcosα＝2si

（1）∵α=π/4,∴c=（√2+sinx,√2+cosx）f（x）=b·c=cosx（√2+sinx）+sinx（cosx+√2）=√2cosx+cosx*sinx*2+√2sinx=-1+（sin

f(x)=向量a×向量b=(sinx,√3cosx)*(cosx,cosx)=sinxcosx+√3cosxcosx=1/2(2sinxcosx+2√3cosxcosx)=1/2(sin2x+√3co

f(x)=sin(x+π/6)cosx+cosxcos(x-π/3)-1/2cos(x-π/3)=cos(π/3-x)=cos[π/2-(x+π/6)]=sin(x+π/6)所以,f(x)=sin(x

f(x)=a*b=sin^x+根号3*cosxsinx=sin平方x+2分之根号3*sin2x

（1）f(x)的最小正周期为π（2）f(x)的值域为[-2,2] 过程如下图： 

[1].cosx+1/3=1,cosx=2/3,x=arxcos2/3.[2].cos^2x=4/9,sin^2x=5/9,sinx=√5/3.cos(2x+π/3)=cos2xcosπ/3-sin2

f(x)=2*sin^2(x)+2sinx*cosx(向量的数量积坐标运算法则）=1-cos2x+sin2x=1+sin2x-cos2x=1+根2*sin(2x-pai/4)-------------

f(x)=2√3cosx^2+2sinxcosx=sin2x+√3(cos2x+1)=sin2x+√3cos2x+√3=2sin(2x+π/3）+√3后面应该会解吧?

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)a⊥b则：f(x)=sinxcosx+√3cosxcosx=sin2x/2+√3(1+cos2x)/2=sin(2x+π/3)+√3/2=0∴2x+π/3=2kπ+3π/2±π/6∴x=kπ+7

a*b=sin²x＋sinxcosx=sinx(sinx＋cosx)=(1/2)(sinx＋cosx),所以,sinx＋cosx=0或sinx=1/2.1、若sinx＋cosx=0,则tan

f(x)=sin(π/2+x)cosx+cos（π-x)sinx=(cosx)^2-cosxsinx=((1+cos2x)/2)-(1/2)sin2x=(1/2)(cos2x-sin2x)+1/2=(