f(x)=2cos(ωx π 6)平移是奇函数,ω最小值

f(x)=cos(-2x+a)(-π

f(x)=cos^2(x-π/6)-sin^2x=（2cos^2(x-π/6)-1+1）/2+(1-2sin^2x-1)/2=[cos(2x-π/3)+coscos2x]/2=(cos2xcosπ/3

由诱导公式f(x)=[6cos(π+x)+5sin^2(π-x)-4]/cos(2π-x)=(-6cosx+5(sinx)^2-4)/cosxf(-x)=[-6cos(-x)+5(sin(-x))^2

f(x)＝cos(ωx−π6)的最小正周期为T=2πw=π5，∴w=10故答案为：10

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）=sinωx+cos（ωx+π6）=sinωx+cosωxcosπ6-sinωxsinπ6=sinωx+32cosωx-12sinωx=12sinωx+32ωx=sin（ωx+π3）∵相邻两条

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

不知道,sorry

解析：f(x)=sin(2x-π/6)+cos²x=sin2x*cos(π/6)-cos2x*sin(π/6)+(cos2x+1)/2=sin2x*√3/2-cos2x*1/2+cos2x*

f(x)=(√3/2sinx/2+1/2cosx/2)(√3/2cosx/2+1/2sinx/2)=1/4(3sinx/2cosx/2+√3sin^2x/2+√3cos^2x/2+cosx/2sinx

再问：详细点嘛！再答：已经很详细了~~

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

f(x)=sin(wx+a)+cos(wx+a)=根号下2sin(wx+a+π/4)最小正周期=2π÷w=πw=2f(x)=根号下2sin(2x+a+π/4)f(x)=根号下2sin(2x+a+π/4

这个用cos（α-β）好想可以做出来,最好问老师

f(X)=sin2x+cos(2x+π/6)=sin2x+（根号3/2)*cos2x-(1/2)*sin2x=(1/2)*sin2x+（根号3/2)*cos2x=sin(2x+π/3）所以,最小正周期

f(x)=√3sinωx+cosωx=2(sinωx*√3/2+cosωx*1/2)=2(sinωx*cosπ/6+cosωx*sinπ/6)=2sin(ωx+π/6)

解:⑴f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+(cosx)^2=-1/2+sinπ/6cos2x-sin2xcosπ/6+cos2xcosπ/3+sin2xsinπ/3+(

f（x）=2*（1-cos2（x+π\4））/2-（根号3/2cos2x-1/2sin2x）=1-sin2x-（根号3/2cos2x-1/2sin2x）=1-（1/2sin2x+根号3/2cos2x）