已知函数f(x)=√3cos²x-2sinxcosx-√3sin²x

先化简,f（x）=sin（2x-π/6）（1）最小正周期T=2π/2=π（2）x∈[0,π/4]2x-π/6∈[-π/6,π/3]∴f（x）单调递增最小值为f（0）=sin（-π/6）=-1/2最小值

就是简单的三角函数,把ωx+φ看成一个整体就行了,强行配成两角和的函数（不熟练可以回去验证看是不是原式）.自然想到（√3）/2和1/2为60度的正余弦值了,就用cos（α+β）=cosαcosβ-si

f(x)=sin²(x)+(√3)sin(x)cos(x)+2cos²(x)=3/2+√3/2sin2x+1/2cos2x=3/2+sin(2x+π/6)函数f(x)的最小正周期T

f(x)=2sin(x/4)cos(x/4)+(√3)cos(x/2)=sin(x/2)+(√3)cos(x/2)=2sin(x/2+π/3)T=4π最大值2最小值-2（2）令g(x)=f(x+π/3

两倍角公式2cos²x=cos2x+12sinxcosx=sin2xf(x)=-4cos²x+4√(3)sinxcosx+5=-2(cos2x+1)+2√3sin2x+5=2√3s

公式好像可以化简为√3sin(2x)+cos(2x)=2[0.5cos(2x)+√3/2sin(2x)]=2[sin60*sin2x+cos60*cos2x]=2sin(60+2x),然后可以继续算下

f(x)=√3sinxcosx+cos²x=(√3/2)sin2x+(1+cos2x)/2-1/2=(√3/2)sin2x+(1/2)cos2x=sin2x*cos(π/6)+cos2x*s

f(x)=cos(2x-π/3)-(cos^2x-sin^2x)=cos(2x-π/3)-cos2x=2sin(2x-π/6)sinπ/6=sin(2x-π/6)因为y=sinx的单减区间为[π/2+

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

已知函数f(x)=(√3)sin(ωx+φ)-cos(ωx+φ)(o0)为偶函数且函数y=f(x)图像的两相邻对称轴间的距离为π/2.求f(π/8)的值;还有一问是：将函数y=f(x)的图像向右平移π

①f（x）=2cos²x+2√3sinxcosx+1=1+cos2x+√3sin2x+1=2sin（2x+π/6）+2T=π单调减区间：2kπ+π/2≤2x+π/6≤2kπ+3π/2（k∈Z

已知函数f(x)=sin(x/2)+(√3)cos(x/2),x∈R；(1)求f(x)的最小正周期,并求函数f(x)在x∈[-2π,2π]上的单调增区间；（2）函数f(x)=sinx(x∈R)的图像经

（1）f（x）=2cosxsinx-2cos²x=sin2x-cos2x-1=√2（√2/2sin2x-√2/2cos2x）-1=√2sin（2x-π/4）-1∴T=π（2）x∈[π/8,3

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

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

f(x)=2√3sinxcosx-2cos(x+π/4)cos(x-π/4)=√3sin2x+2sin(x+π/4-π/2)cos(x-π/4)=√3sin2x+sin(2x-π/2)=√3sin2x

f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]=4cos(x+π/3)[1/2sin(x+π/3)-√3/2cos(x+π/3)]=4cos(x+π/3)[sin(

f(x)=(cosx)^4-(sinX)^4+2√3sinxcosx=[(cosx)^2+(sinx)^2][(cosx)^2-(sinx)^2]+√3sin2x=(cosx)^2-(sinx)^2+

解f(x)=sinx+√3cosx=2sin(x+π/3)∴最小正周期为：T=2π/1=2π(2)当π/2+2kπ

f(x)=sinx-√3cosx+1=2sin(x-π/3)+11.2kπ-π/2