(I)求函数f(x)的最小正周期和图像的对称轴方程
(II)求函数f(x)在区间[-π/12,π/2]上的值域
已知函数f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4)
(1)函数f(x)的最大值及最大值的自变量x的集合;(2)函数f(x)的单增区间;解:(1)f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4)=cos(2x-π\/3)+cos(-π\/2)-cos(2x)=cos2xcos(π\/3)+sin2xsin(π\/3)-cos2x=(1\/2)cos2x+(√3\/2)sin2x-cos2x =(√3\/2)sin2x-(1\/2)...
已知函数f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4) 求函数f(x
sin(x+π\/4)=sin(x-π\/4+π\/2)=cos(x-π\/4)所以原式=cos(2x-π\/3)+2sin(x-π\/4)cos(x-π\/4)=cos(2x-π\/3)+sin(2x-π\/2)=1\/2cos(2x)+√3\/2sin(2x)-cos(2x)=√3\/2sin(2x)-1\/2cos(2x)=-cos(2x+60)最小正周期=2π\/2=π 对称轴=kπ\/2 根据图像和单...
已知函数f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4),求最小正周期
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4)=cos(2x-π/3)-cos[(x-π/4)+(x+π/4)]+cos[(x-π/4)-(x+π/4)]=cos(2x-π/3)-cos2x T=2π\/2=π
已知函数为f(x)=cos(2x-π╱3)+2sin(x-π\/4)sin(x+π\/4)
解:f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/2-π\/4)=cos(2x-π\/3)+2sin(x-π\/4)cos(x-π\/4)=cos(2x-π\/3)+sin(2x-π\/2)=cos(2x-π\/3)-cos2x =√3\/2sin2x-1\/2cos2x =sin(2x-π\/6)最小正周期T=π,对称轴方程x=kπ\/2+5\/12π(k为整数)(2)-π\/...
已知函数f(x)=cos(2x-派\/3)+2sin(x-派\/4)sin(x+派\/4)(1)求函数f(x...
解:f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4)=cos2xcosπ\/3+sin2sinπ\/3+2(sinxcosπ\/4-cosx-sinπ\/4)(sinxcosπ\/4-cosx-sinπ\/4)=cos2x\/2+√3\/2sin2x+sin²x-cos²x =cos2x\/2+√3\/2sin2x-cos2x =-cos2x\/2+√3\/2sin2x =cos(2x+2π\/3)∴T=...
已知函数fx=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4)
f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4)=cos(2x-π\/3)+2sin(x-π\/4)cos[π\/2-(x+π\/4)]=cos(2x-π\/3)+2sin(x-π\/4)cos(π\/4-x)=cos(2x-π\/3)+2sin(x-π\/4)cos(x-π\/4)=cos(2x-π\/3)+sin(2x-π\/2)=cos2x*(1\/2)+sin2x*(√3\/2)-cos2x =(...
已知旱数fx=cos(2x-派\/3)+2sin(x-派\/4)sin(x+派\/4)
f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4)=cos(2x-π\/3)+[cos(x-π\/4-x-π\/4)-cos(x-π\/4+x+π\/4)]=cos(2x-π\/3)+cos(-π\/2)-cos(2x)=cos2x*1\/2+sin2x*根号3\/2-cos2x =sin(2x-π\/6)∈[-1,1]
已知函数f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4)
f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4)=1\/2cos 2x+根号3\/2sin 2x+2*(根号2\/2sin x-根号2\/2cos x)(根号2\/2sin x+根号2\/2cos x)=1\/2cos 2x+根号3\/2sin 2x+ 2*根号2\/2 * 根号2\/2 *(sin x+cos x)(sin x-cos x)=1\/2cos 2x+根号3\/2sin 2x+ ...
已知函数f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4),问:求函数f(x...
2x)=√3sin(2x-π\/3)1. 可见,f(x)最小正周期为2π\/2=π 图像为g(x)=√3sin(2x)向右平移π\/6个单位得到 则对称轴为,x=kπ\/2+5π\/12(k∈Z)2. 令z=2x-π\/3 由x∈[-π\/12,π\/2]则z∈[-π\/2,2π\/3]√3sin z∈[-√3,√3]即f(x)值域为[-√3,√3]
已知函数f(x)=cos(2x-π\/3)+2sin(x-π\/4)sin(x+π\/4)
2sin(x-π\/4)sin(x+π\/4)=cos(x-π\/4-x-π\/4)-cos(x-π\/4+x+π\/4)=-cos2x f(x)=cos(2x-π\/3)-cos2x=cos(2x-π\/6-π\/6)-cos(2x-π\/6+π\/6)=2sin(2x-π\/6)sin(π\/6)=sin(2x-π\/6)最小正周期是2π\/2=π,图象对称轴很容易知道是π\/3+nπ\/2 ...