函数f(x)=cos(πx φ)的部分图像如图所示

∵f（x）＝cos（2x－π/3）＋（sinx）^2－（cosx）^2＝cos（2x－π/3）－cos2x＝2sin（2x－π/6）sin（π/6）＝sin（2x－π/6）.∴g（x）＝［sin（2x

1f(x)=√3sinπx+cosπx=2（（√3/2）sinπx+（1/2）cosπx）=2sin（πx+π/3）∴最小正周期T=2π/w=2π/π=2值域f（x）∈[-2,2]2-π/2+2kπ＜

f(x)=cos(x-π/)+sin^2x-cos^2x=-cosx+sin^2x-cos^2x=-2cos^2x-cosx+1最小正周期2π

(1)f(x)=cos(x+2π/3)+2cos²(x/2)=-(cosx)/2-(√3sinx)/2+1+cosx=1-[(√3sinx)/2-(cosx)/2]=1-[sin(x-π/6

1)f(x)=1+cos(2x+π/3)-(1+cos2x)/2=1/2-sin2x根号3/2最小值1/2-根号3/2最小正周期π2）c带入得sinC=根号3/2C=π/3A=π-B-C=2π/3-a

根据公式：COS^2a=(1+COS2a)/2a=(X+π/12)说句不好听的,你还是基本知识没掌握好,不会活用知识.希望你多背多看,看清题,把公式活用.

f(x)=-√2sin(2x+π/4)+6sinxcosx-2cos²x+1=-√2(sin2xcosπ/4+cos2xsinπ/4)+3sin2x-2×(1+cos2x)/2+1=-√2(

f(x)=cos(2π-x)-x³sin1/2x=cosx-x³sin1/2x函数定义域为Rf(-x)=cos(-x)-(-x)³sin(-1/2x)=cosx-x

1、f(x)=cos(2π-x)cos(π/2-x)-sin^2x=-cosxsinx-sin^2x=-½sin2x-(1-cos2x)/2=-1/2sin2x+1/2cos2x-1/2

利用二倍角公式,则有：f(x)=-2(sin2x)^2/(-2sin2x)+cos2x=sin2x+cos2x=根号2sin(2x+pi/4)所以最小正周期为pi,2kpi+pi/2

f(x)=cos(-x/2)+sin(π-x/2)=cosx/2+sinx/2f(a)=cos(a/2)+sin(a/2)=（2√10）/5cos(a/2)+sin(a/2)=（2√10）/5平方1+

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+

(^2)x这是什么啊完全看不懂诶.再问：就是（sinx）^2再答：啊啊懂啦再答：

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

F（0）=0cos(φ-π/6)=0φ-π/6=π/2+kπ又0

f(x)=sin(2x)+√2cos(x-π/4)=sin(2x)+√2[cosxcos(π/4)+sinxsin(π/4)]=sin(2x)+cosx+sinx=sin(2x)+√2sin(x+π/

f(π/3-x)=cos(π/3-x)/cos(π/6-(π/3-x))=cos(π/3-x)/cos(x-π/6)=cos(π/3-x)/cos(π/6-x)注：cosx=cos(-x)所以：f(x

∵函数f（x）=2cos（π4-ωx）=cos（ωx-π4）（ω＞0）的最小正周期为π2，∴2πω=π2，ω=2，∴f（x）=cos（2x-π4）．令2kπ≤2x-π4≤2kπ+π，k∈z，求得kπ+

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(

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