点X=0是函数f(x)=cosx x的

(x)=cos^4x+2sinxcosx-sin^4x=cos^2x-sin^2x+2sinxcosx=cos2x+sin2x=√2sin(2x+π/4)在[0,π//2]f(x)最小值=1,x=0

你都没x再问：当x0时f(x)=cosπx与f(x)=log3(x)的交点个数。画图像，看一下有三个交点所以有三对再问：画来再答：再问：关于y轴对称的点共有几对

f(x)=cos²x-2cos²（x/2)f'(x)=-2sinxcosx+sinx=sinx(1-2cosx)单调增sinx>0,cosx

f（x）=cos^2x-2cos^2x/2=（2cos^2x/2-1)^2-2cos^2x/2(倍角公式）设：cos^2x/2=m(1≥m≥0）则：f（x）=（2m-1)^2-2m=4m^2-6m+1

f(x)=-cos2x所以最小正周期为2π/2=π注意：求最小正周期的时候一定要将式子化成最简.

f(x)=(1-cos2x)/2+3(1+cos2x)/2=2+cos2x,T=2π/2=π,最小正周期为π.

cos^2(x+∏/3)+cos^2(x-∏/3)=(cosx/2-根号3*sinx/2)^2+(cosx/2+根号3*sinx/2)^2=(cosx)^2/2+3(sinx)^2/2=1/2+(si

0+2kπ再问：0+2kπ

只要cos(sinx)≥0就行了而sinx∈[-1,1]即-1弧度到1弧度在这个范围内余弦值始终为正所以定义域是R

f'(x)={(cosx)'x-x'cosx}/x²=-{xsinx+cosx}/x²再问：为什么是+cosx呢再答：(cosx)'=-sinx然后把负号提出来了再问：呵呵，没看见

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)=sin(x+α)-2cos(x-α)是奇函数f(-x)=-f(x)sin(-x+a)-2cos(-x-a)=-(sin(x+a)-2cos(x-a))-sin(x-a)-2cos(x+a)=

f(x)=cos(x+π/6)-cos(x-π/6)+SQR（3）*cosx=cosxcosπ/6-sinxsinπ/6-cosxcosπ/6-sinxsinπ/6+SQR（3）*cosx=-sinx

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

f(x)=2(1+cosx)/2+cosxcosπ/3-sinxsinπ/3=-(√3/2*sinx-3/2*cosx)+1=-3/2*sin(x-π/3)+1所以T=2π/2=π

(1)f(x)=sinwxcoswx+coswxcoswx=1/2sin2wx+1/2cos2wx+1/2=√(根号)2/2sin(2wx+π/4)+1/2因为f（x）的周期为π,所以w=1f(x)=

f(x)=2cosx*sinx-2cosx^2+1f(x)=sin2x-cos2xf(x)=根号2*sin(2x-45)周期T=π

f(x)=[cos(x)+sin(x)]sin(x)=cos(x)sin(x)+sin^2(x)=1/2sin2x+1/2(1-cos2x)=√2/2[cos(p/4)sin2x-sin(p/4)co

0再问：确定对不?要是对就直接满意答案~~再答：200%肯定

1.cos(x+60)=cos60cosx-sin60sinx=0.5cosx-(√3/2)sinxcos(60-x)=cos60cosx+sin60sinx=0.5cosx+(√3/2)sinx∴f