已知函数f(x)=cos^2(x/2)-sin(x/2)*cos(x/2)-1/2求最小正周期和值域,当f=(α)=3*√2/2,求sin2αcos^2(x/2)=(1+cosx)/2,不是这样么
问题描述：
已知函数f(x)=cos^2(x/2)-sin(x/2)*cos(x/2)-1/2
求最小正周期和值域,当f=(α)=3*√2/2,求sin2α
cos^2(x/2)=(1+cosx)/2,不是这样么
答
f(x)=cos^2(x/2)-sin(x/2)*cos(x/2)-1/2
=1/2(cosx-sinx)
=√2/2cos(x+π/4)
最小正周期：[-π/4,7π/4]
值域：[-√2/2,√2/2]
√2/2cos(x+π/4)=3√2/2
cos(x+π/4)=3(不成立)
答
f(x)=cos²(x/2)-sin(x/2)*cos(x/2)-1/2=1/2*[2cos²(x/2)-1]-1/2*sinx=1/2*cosx-1/2*sinx=√2/2*(√2/2*cosx-√2/2*sinx)=√2/2*cos(x+π/4)最小正周期为：T=2π/1=2π∵-1≤cos(x+π/4)≤1∴-√2/2≤f(x)...