设fx等于2根号3sin(π减x)

f(x)=[(cosx)^2-(sinx)^2]+√3sin2x=cos2x+√3sin2x=2sin(2x+π/6),最小正周期T=π,由-π/2+2kπ≤2x+π/6≤π/2+2kπ,k∈Z解得：

f(x)=√3sin²x+sinxcosx=√3[(1-cos2x)/2]+1/2sin2x=1/2sin2x-√3/2cos2x+√3/2=sin(2x-π/3)+√3/2∵x∈[π/2,

设函数fx=sin(φ-2x)(0

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

0因为根号≥0这是定理

你好,这题应该这样1.f(x)=cos(2x+π/3)+sin²X=负二分之根号三sin2x+二分之一所以最大值为﹙√3+1﹚／2最小正周期为π2.可知COSB=1/3sinC=√3／2∵C

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

fx=2cos^2x+2根号3sinxcosx-1=2cos^2x-1+2根号3sinxcosx根据倍角公式,sin2α=2sinαcosαcos2α=2cos^2（α）-1fx=cos2x+根号3s

再问：第5步为什么要提出一个√3/3，sin前面的1/2去哪了？再答：1/2哪去了？哪也没去啊？只是换了一种存在的方式而已：[(√3)/3]×[(√3)/2]=1/2

 再答：这是高一的题目吧再答：不谢，复习加油

F（X）=cos(√3x+t）F'（X）=-√3sin(√3x+t）F（X）+F'（X）=cos(√3x+t）-√3sin(√3x+t）是奇函数所以F（0）+F'（0）=0即cost-√3sint=0

答：f(x)=2sin(x-π/3)cosx+sinxcosx+√3(sinx)^2=sin(x-π/3+x)+sin(x-π/3-x)+sinxcosx+(√3/2)(1-cos2x)=sin(2x

答：y=f(x)=2√3sinxcosx-2sin²x=√3sin2x+cos2x-1=2*[(√3/2)sin2x+(1/2)cos2x]-1=2sin(2x+π/6)-1y=f(x)关于

化简得到f(x)=2sin(2x-%pi/6)+1x属于(0,2%pi/3),所以f(x)属于（0,3]已知M-2

设函数fx=2cos^2(π/4-x)+sin(2x+π/3)-1=cos(PI/2-2x)+sin(2x+PI/3)=sin(2x)+sin(2x)/2+cos(2x)*sqrt(3)/2=sqrt

（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

解答；f(x)=sin(2x+3分之π)∴sin(2x+π/3)=-3/5∵x∈(0,π/2)∴2x+π/3∈(π/3,4π/3)∵sin(2x+π/3)

解1当2kπ-π/2≤2x+π/3≤2kπ+π/2,k属于Z时,y是增函数即2kπ-5π/6≤2x≤2kπ+π/6,k属于Z时,y是增函数即kπ-5π/12≤x≤kπ+π/12,k属于Z时,y是增函数

f(x)=sin(x/2)cos(x/2)+√3*sin²(x/2)+√3/2=1/2*sinx+√3/2*(1-cosx)+√3/2=1/2*sinx-√3/2*cosx+√3=sin(x

f(x)=-√2sin(2x+∏/4)+2cosx(3sinx-cosx)=-(sin2x+cos2x)+6sinxcosx-2(cosx)^2=-sin2x-cos2x+3sin2x-(1+cos2