f(x)=2sin平方x 2根号3sinxcosx 1

作参考

f(x)=2sin平方x-根号3sinxcosx+cos平方x=1-cos2x-√3/2sin2x+1/2(cos2x+1)=1-cos2x-√3/2sin2x+1/2cos2x+1/2=3/2-1/

因为sin^2x=（1-cos2x）/2所以f(X)=sin(2x-π/4)-2根号2sin^2x=√2/2sin2x-√2/2cos2x-√2(1-cos2x)=√2/2sin2x+√2/2cos2

我刚答过的题:(I)f(x)=3sin^2x+2√3sinxcosx+5cos^2x=sin^2x+2√3sinxcosx+3cos^2x+2(sin^2x+cos^2x)=sin^2x+2√3sin

f(x)=(sin²x+cos²x)+(√3/2)(2sinxcosx)+cos²x=1+(√3/2)sin2x+(1+cos2x)/2=(√3/2)sin2x+(1/2

f(x)=sin平方x+2倍根号3sinxcosx+3cos平方xf(x)=1+√3sin2x+3/2+3/2*cos2x=5/2+√3sin2x+3/2cos2x=5/2+√21/2sin(2x+φ

函数f(x)=2sin^2(x+π/4)-√3cos2x-1=-cos(2x+π/2)-√3cos2x=sin2x-√3cos2x=2sin(2x-π/3)1.当x属于R时,函数f(x)的最小正周期T

f（x）=2cosx*sin（x+π/3)-√3sinx^2+sinx*cosx=2cosx*(sinxcosπ/3+cosxsinπ/3))-√3sinx^2+sinx*cosx=sinxcosx+

f(x)=-根号3cos2X-sin2x=-2(根号3/2cos2x+1/2sin2x)=-2sin(2x+π/3)(1).T=2π/w=π（2）.由X属于[-π/3,π/3]得2x+π/3∈【-π/

f(x)=√3/2sin2x-1/2(1-cos2x)+1/2=√3/2sin2x-1/2+1/2cos2x+1/2=√3/2sin2x+1/2cos2x=sin(2x+π/3)

f(x)=√(x²+2x+2)+√(x²-4x+8)=√[(x+1)²+1]+√[(x-2)²+4].分析一,√[(x+1)²+1]取最小值是1时,√

y=sinx^2+根3sinxcosx+2cosx^2=-1/2(1-2sinx^2)+1/2根3*2sinxcosx+2cosx^2-1+3/2=-1/2cos2x+二分之根3倍sin2x+cos2

因为f(x)=根号3sin(2x-π/6)+2sin的平方（x-π/12)=根号3sin(2x-π/6)-（1-2sin的平方（x-π/12)）+1=根号3sin(2x-π/6)-cos(2x-π/6

f(x)=2√3sin²x-sin(2x-π/3)=√3-√3cos2x-1/2sin2x+√3/2cos2x=√3-(1/2sin2x+√3/2cos2x)=√3-sin(2x+π/3)T

f(x)=3sin平方x+2根号3sinxcosx-3cos平方x=根号3sin2x-3cos2x=2根号3sin(2x-π/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

前提掌握：sinx*sinx+cosx*cosx=1cos2x=2*cosx*cosx-1=1-2*sinx*sinxcos（x-π/4）=-sin（x-π/4+π/2）=-sin(x+π/4)sin

(1)已知函数f(x)=-(√3)sin²x+sinxcosx,求f(25π/6).f(x)=-(√3)sin²x+sinxcosx=(√3/2)(cos2x-1)+(1/2)si

f(x)=sin²x+√3sinx*sin(x+π/2)=sin²x+√3sinxcosx=(1/2)(1-cos2x)+(√3/2)sin2x=(√3/2)sin2x-(1/2)

f(x)=(1-sqrt(3))cos2x+1+sqrt(3)最大值：2；最小值：-sqrt（3）周期：πsqrt(3)表示根号3-------------------改后你的补充表述及解答仍有问题,