в)
Проверяем равенство:
f(x)=f(x+T)
[m]\frac{1}{3}ctg(\frac{x}{3}-\frac{π}{2})+1=\frac{1}{3}ctg(\frac{x}{3}-\frac{π}{2}+\frac{π}{3})+1[/m]
[m]\frac{1}{3}ctg(\frac{x}{3}-\frac{π}{2})=\frac{1}{3}ctg(\frac{x}{3}-\frac{π}{2}+\frac{π}{3})[/m]
[m]ctg(\frac{x}{3}-\frac{π}{2})=ctg(\frac{x}{3}-\frac{π}{2}+\frac{π}{3})[/m]
[m]ctg(\frac{x}{3}-\frac{π}{2})=ctg(\frac{x}{3}-\frac{π}{6})[/m]
[m]ctg(\frac{x}{3}-\frac{π}{2})-ctg(\frac{x}{3}-\frac{π}{6})=0[/m]
[m]\frac{cos(\frac{x}{3}-\frac{π}{2})}{sin(\frac{x}{3}-\frac{π}{2})}-\frac{cos(\frac{x}{3}-\frac{π}{6})}{sin(\frac{x}{3}-\frac{π}{6})}=0[/m]
[m]\frac{cos(\frac{x}{3}-\frac{π}{2})\cdot sin(\frac{x}{3}-\frac{π}{6})-sin(\frac{x}{3}-\frac{π}{2})\cdot cos(\frac{x}{3}-\frac{π}{6}) }{sin(\frac{x}{3}-\frac{π}{2})\cdot sin(\frac{x}{3}-\frac{π}{6})}=0[/m]
В числителе формула синуса разности:
[m]\frac{ sin(\frac{x}{3}-\frac{π}{6}-(\frac{x}{3}-\frac{π}{2}))}{sin(\frac{x}{3}-\frac{π}{2})\cdot sin(\frac{x}{3}-\frac{π}{6})}=0[/m]
[m]\frac{ sin(\frac{x}{3}-\frac{π}{6}-\frac{x}{3}+\frac{π}{2})}{sin(\frac{x}{3}-\frac{π}{2})\cdot sin(\frac{x}{3}-\frac{π}{6})}=0[/m]
[m]\frac{ sin\frac{π}{3}}{sin(\frac{x}{3}-\frac{π}{2})\cdot sin(\frac{x}{3}-\frac{π}{6})}=0[/m]- неверно,
г)
Проверяем равенство:
f(x)=f(x+T)
[m]tg5x+2,5=tg5(x+\frac{π}{5})+2,5[/m]
[m]tg5x=tg(5x+π)[/m] - верно по формулам приведения.