函数f(x)=2cos(x π 3)-1的对称轴为

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函数f(x)=2cos(x π 3)-1的对称轴为
设函数f(x)=cos(2x+π/3)+sin方x

f(x)=cos(2x+π/3)+sin²x=cos2xcosπ/3-sin2xsinπ/3+[1-cos(2x)]/2=1/2cos2x-√3/2sin2x+1/2-1/2cos2x=-√

已知函数f(x)=cos(2x-π/3)+sin^2x-cos^2x,设函数g(x)=[f(x)]^2+f(x),求g(

∵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

设函数f(x)=cos(x+2/3π)+2cos^2 x/2,x∈R.

(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

设函数f(x)=2cos^2(x+π/6)-cos^2x

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

已知函数f(x)=cos^2(x+π/12).

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.

1.设函数f(x)=cos(2x+π/3)+sin(平方)x

1.(1)f(x)=cos(2x+π/3)+sin(平方)x=1/2cos2x-根号3/2sin2x+sin(平方)x+1/2-1/2=1/2cos2x-根号3/2sin2x-1/2cos2x+1/2

设函数f(x)=cos(2x+π/3)+sin^2 X

f(x)=cos2x*1/2-√3*sin2x+(1-cos2x)/2=cos2x-√3sin2x+1/2=2cos(2x+π/3)+1/2所以最小正周期T=2π/2=π当cos(2x+π/3)=1取

设函数f(x)=cos(2x+π/3)+sin^2x

f(x)=cos(2x+π/3)+sin^2X=1/2cos2x-根号3/2sin2x+(1-cos2x)/2=1/2-根号3/2sin2x因为f(c/2)=-1/4,所以sinC=根号3/2,cos

设函数f(x)=cos(2x+π/3)+sin²x

(1)f(x)=cos(2x+π/3)+sin²x=1/2cos2x*-√3/2sin2x*+(1-cos2x)/2=1/2-√3/2*sin2xT=2pi/2=pi最大值是1/2+√3/2

设函数f(x)=cos(2x+π/3)+sin^2x-1/2

f(x)=cos(2x+π/3)+sin^2x-1/2=cos(2x+π/3)+(1-cos2x)/2-1/2=cos2xcos(π/3)-sin2xsin(π/3)-cos2x*1/2=-√3/2*

急.设函数f(x)=cos(2x+π/3)+sin^2 X

原式=1/2+根3/2sin2X1)求函数f(x)的最大值1/2+根3/2,最小正周期π

设函数f(x)=cos(2x+π/3)+sin²X

f(x)=cos(2x+π/3)+sin²X=1/2*cos2x-√3/2*sin2x+(1/2)(1-cos2x)=1/2-√3/2*sin2x,(1)f(x)的最大值=(1+√3)/2.

若函数f(x)=[2cos^3 x-sin^2 (x+π)-2cos (-x-π)+1]/[2+2cos^2(7π+x)

化简,f(x)=cosx(1)不用证了吧(2)1/2再问:过程再答:先将所有的带π的全部化成标准的“-sin^2(x+π)”化成“-sin^2(x)”“-2cos(-x-π)”化成“2cosx”“2c

证明f(x)=cos^2x+cos^2(x+∏/3)+cos^2(x-∏/3)是常数函数

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

函数f(x)=cos(2π-x)-x^3·sinx的奇偶性为

是偶函数  图像作参考

已知函数f(x)=cos(2x-π/3)+sin^2 x-cos^2 x

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+

已知函数f(x)=cos(2x-π/3)+sin(^2)x+cos(^2)x.求化简~

(^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:

已知函数f(x)=cos(2x-π\3)+sin²x-cos²x

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

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(