求函数sin²x 根号3cosx 1 4

f(x)=2cosx*sin(x+π/3)-根号3sinx^2+sinxcosx=2cosx*(1/2sinx+根号3/2cosx)-根号3sinx^2+sinxcosx=2sinxcosx+根号3c

因为：函数f(x)=2cosx·sin（x+π/3)-√3(sinx)^2+sinx·cosx=2cosx·sin(x+∏/3)-√3(sinx)^2+sinx·cosx=2cosx·sinx·cos

1.y=√2sin(x+π/4),所以值域是[-√2,√2]2.y=1-cos^2x-cosx+1=-cos^2x-cosx+2设t=cosx,y=-t^2-t+2(=-1

y=f(x)=(sinx)平方-根号3sin(pai+x)*cosx=(1/2)(1-cos2x)+根号3sinxcosx=(√3/2)sin2x-(1/2)cos2x+1/2=sin(2x-π/6)

f(x)=cos(x/2)=√3sin(x/2)cos(x/2)+cos^2(x/2)=√3/2sinx+1/2(cosx+1)=√3/2sinx+1/2cosx+1=sinxcos(π/6)+cos

f(x)=cos^2x+√3sinx*cosx=(cos2x+1)/2+√3/2sin2x=√3/2sin2x+1/2cos2x+1/2=sin(2x+π/6)+1/2所以当2x+π/6=2kπ+π/

答案：cosx=-1或cosx=1/4根据和差化积公式,[sin(x+120°)-sin(x-120°)]/sin(x+120°)*sin(x-120°)=4根号3/3,(2cosxsin120°)/

y=2sin²x+2√3sinxcosx-2=1-cos(2x)+√3sin(2x)-2=2sin(2x-π/6)-1最小正周期=2π/2=π当sin(2x-π/6)=1时,y有最大值yma

f(x)=[2sin(x+π/3)+sinx]cosx-根号3sin^2x和差角公式展开f(x)=(sinx+根号3cosx+sinx)cosx-根号3sin2x=2sinx*cosx+根号3cosx

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

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

这题目是不是Y=SIn2X2SinX=cosx+3cosX?感觉不对劲啊请完善题目

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：

y=根号3sinx+cosx+2012=2(根号3/2sinx+1/2cosx)+2012=2sin(x+π/6)+2012所以周期为2π最大值2014最小值2010

f(x)=sin(2x+π/3)(积化和差公式）所以最小正周期T=2π/2=π

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

最小正周期为“派”,原=2cx(1/2sx+根3/2cx)-根3/2=cxsx+根3/2(cx)^2-根3/2=1/2s(2x)+根3/2c(2x)=s(2x+派/3)；所以T=2派/2=派注:sx:

（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