a=(2根号3sinx,sinx cosx)

sin^3X+cos^3X=（sinx+cosx)(sin^2x-sinxcosx+cos^2x)=根号2(1+sinxcosx)sinX+cosX=根号2所以sin^2x+cos^2x+2sinxc

2sin(X+π/3)=2(sinxcosπ/3+cosxsinπ/3)=2(1/2*sinx+二分之根号3*cosx)=Sinx+(根号3)*Cosx

17/4根号3/2-321/64a/2带入后,平方,化简

f(x)=√3sinx+sin(π/2+x)=√3sinx+cosx=2sin(x+π/6)∴最大值是2

f(x)=2sin^2-2√3*sinX*cosX+1π=1-cos2X-√3*sin2X+1=2-2sin(2X+π/3)T=2π/2=π

√2sin(x-π/4)=√2(sinxcosπ/4-cosxsinπ/4)=√2[(1/√2)sinx-(1/√2)cosx]=sinx-cosx

因为sin(π/6)=1/2；cos(π/6)=(√3)/2所以原式等于cos(π/6)sinx+sin(π/6)cosx利用两角和公式sin(A+B)=sinAcosB+cosAsinB得到sin（

[sinx+cos(π+x)]/[sinx+sin(π/2-x)]=(sinx-cosx)/(sinx+cosx)=(tanx-1)/(tanx+1)=1-[2/(tanx+1)]sinx=√3/3∴

f(x)=2根号3*sinx/3*cosx/3-2sin平方x/3=4sinx/3﹙√3cosx/3﹣sinx/3﹚/2=4sinx/3sin﹙π﹣x﹚/3=﹣2[cosπ/3－cos﹙2x-π﹚/3

1、f(x)=sin²x+√3sinxcosx=(1-cos2x)/2+√3sin2x/2=(√3/2)sin2x-(1/2)cos2x+1=sin(2x-π/6)+1/2∴当x∈{x|(2

这个简单：f(x)=2cosx(sinxcos(pi/3)+cosxsin(pi/3))-根号33sin^2x+sinx*cosx=2sinxcosx+根号3cos2x=2sin(x+pi/3)所以：

f(x)=sinx+sin(x+60°)=sinx+sinx/2+√3cosx/2=3sinx/2+√3cosx/2=√3sin(x+30°)故f(x)max=√3再问：sinx+sinx/2为什么等

（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： 

f(x)=[2sin(x+π/3)+sinx]cosx-√3sin²x=[2sinxcos(π/3)+2cosxsin(π/3)+sinx]cosx-√3sin²x=(2sinx+

3sinx-2sin^2x-1>=0令sinx=tt范围为[-1,1]3t-2t^2-1>=0所以(t-1/2)(t-1)

（1）化简可得f(x)=(sin(x/2))^2+((√3)/2)sinx-0.5f'(x)=sin(x/2)cos(x/2)+((√3)/2)cosx=sinx+√3cosx=0√3cosx=-si

（1）、f(x)的定义域为sinx≠0,即x≠kπ；如果定义x=kπ时f(x)等于-1,可将定义域扩大至整个实数域；f(x)=[2√3sinxcosx-2cos²x]+1=√3sin2x-c

f(x)=[2sin(x+π/3)+sinx]cosx-√3sin²x=[sinx+√3cosx+sinx]cosx-√3sin²x=2sinxcosx+√3cos²x-

想成整理后得f(x)=sin2x+√3cos2x=2sin(2x+3/π),周期为π单调增区间,2x+3/π∈[-π/2+2kπ,π/2+2kπ],x∈[-5π/12+kπ,π/12+kπ]单调减区间