已知函数fs等于-根号2sin

2kπ+π/4或者是2kπ+3π/4

sina+cosa=√2(√2/2sina+√2/2cosa)=√2(sinacosπ/4+cosasinπ/4)=√2sin(a+π/4)-1

数理答疑团为您解答,希望对你有所帮助.Sinθ+Cosθ=√2,(Sinθ+Cosθ)²=2,Sin²θ+Cos²θ+2SinθCosθ=2,SinθCosθ=1/2;(

f(x)=cos^2x-sin^2x+2(根号3)sinxcosx+1=cos2x+(根号3)sin2x+1=2{(1/2)cos2x+[(根号3)/2]sin2x}+1=2sin(2x+派/6)+1

是四份之π还是π分之四?再问：四分之π再答：你的题给的有点特殊，所以只能按照我理解的给答案，如果有错可以追问f(x)=ab={√2sin(π/4+x)+1}x{√2sin(π/4+x)-1}-√3co

f(x)=√3(sin^2x-cos^2x)-2sinxcosx=-√3cos2x-sin2x=-2sin(2x+π/3)1.求最小正周期T=π2.设x∈[-π/3,π/3],求函数的值域和单调区间-

(1)因为sin²x=（1-cos2x）/2,sin(2x-π/3)=1/2sin2x-根号3/2cos2x所以函数f(x)=2倍根号3sin²x-sin(2x-π/3)=-sin

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

f(x)=2根号3sin方x+sin2x+根号3＝根号3（2sin方x＋1）＋sin2x＝根号3（1－cos2x＋1）＋sin2x＝2根号3－根号3cos2x＋sin2x＝2sin（2x－60度）＋2

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

通分即可1/1+sinα+1/1-sinα=(1-sinα)/[(1+sinα)(1-sinα)]+(1+sinα)/[(1-sinα)(1+cosα)]=(1-sinα)/(1-sin²α

经济将积极

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：

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

1：(sinwx)^2+√3sinwxsin(wx+π\2)=(sinwx)^2+√3sinwxcoswx=2[(sinwx)^2+(√3\2)sin2wx]\2=[2(sinwx)^2+√3sin2

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

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

1、∵1-cos2x=2sin^2∴f(x)=2sinxsinx+2√3sinxcosx+1=1-cos2x+√3sin2x+1=2+2(√3/2*sin2x-1/2*cos2x)=2(cos(π/6

1、f(x)=2sinxsinx+2√3sinxcosx+1=1-cos2x+√3sin2x+1=2sin(2x-π/6)+2对称轴：2x-π/6=kπ+π/2,得x=kπ/2+π/3对称中心：2x-