一道高一数学题（求详细过程）8

f(x)=sin(ax+π)=-sin(ax)

设最小正周期T
f(x)=f(x+T)
-sin(ax)=-sin(ax+aT)
-sin(ax)=-sin(ax)cos(aT)-cos(ax)sin(aT)
sin(ax)[cos(aT)-1]+cos(ax)sin(aT)=0
此式恒成立
则cos(aT)-1=0 sin(aT)=0
即得aT=2kπ
T=2kπ/a

最小正周期|2π/a|

f(x+|2π/a|)=sin(ax+a*|2π/a|+π)=sin(ax+π)=f(x)

f(x+|π/a|)=sin(ax+a*|π/a|+π)
当a>0时
f(x+|π/a|)=sin(ax+2π)=-f(x)
当a<0时
f(x+|π/a|)=sin(ax)=-f(x)
可见|π/a|不是最小正周期