2根号3cos平方x-2sinxcosx-根号3位移T奇函数

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/

原式=1/2sinx+根号3/2cosx+sinx-根号3cosx+根号3/2cosx-3/2sinx（先全部展开）=0（最后合并同类项得0）

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

1+cos²x-sin²x>02cos²x>0cos²x>0即cos²x≠0x≠kπ+π/2,k∈Z定义域为{x∈R|x≠kπ+π/2,k∈Z}再问：

f(x)=(sinx)^2+2√3sin(x+π/4)cos(x-π/4)-(cosx)^2-√3=-((cosx)^2-(sinx)^2)+√3(sin(x+π/4+x-π/4)+sin(x+π/4

根号3（根号3/2*cosx-1/2sinx）=根号3（sin60cosx-cos60sinx)=根号3*sin(60-x)这里注意将根号3提取出来.为什么提出根号3呢因为根号下【(3/2）^2+((

12190394=49302/4565*64565-6再问：我去年买了个大金表

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/

先来第一题,一会发另外一半第二题来了：解答完毕,纯手打截图,望采纳,祝学习愉快!~

f(x)=2cosx·sin(x+π/3）-√3sin²x+sinx·cosx=2cosx·（sinx·cosπ/3+sinπ/3·cosx）-√3sin²x+sinx·cosx=

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

请问你的题目是否有错误?感觉应该是f(x)=2√3sin(π-x)cosx+2cos的平方x-1.f(x)=2√3sin(π-x)cosx+2cos的平方x-1.=2√3sinxcosx+2cosx^

1.√3sinx+cosx=2*(sinx*√3/2+cosx/2)=2(sinxcosπ/6+cosxsinπ/6)=2sin(x+π/6)2.用万能公式可得字数受限

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）

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

原式=sin（x+π/3）+√3cos（x+π/3）+2sin（x-π/3）=2[1/2sin(x+π/3)+√3/2cos(x+π/3)]+2sin(x-π/3)=2sin(x+π/3+π/6)+2

3/2cosx-3/2(sinx)^2

f(x)=(sinx)^2+「3sinxcosx+2(cosx)^2=(sinx)^2+(cosx)^2+(cosx)^2+「3sinxcosx=1+(1+cos2x)/2+(「3/2)*2cosxs