[Math] Period T of function $y = \cos(nx)$

Question

There are two functions:

1) $f(x) = \cos(nx)$

2) $f(x) = \cos(x)$

$T=2 \pi$ is the fundamental period of $(2)$ function.

$T_1$ is the fundamental period of $(1)$ function.

How to prove that $T_1=\frac{2\pi}{n}$?

Answer 1

Use the definition of periodic functions, let $T_1$(the smallest non negative real) be the period of the first function so:

$$ f(x) = f(x+T_1)\quad (T_1\neq0)\\ \cos{nx} = \cos{n(x+T_1)}\\ \cos{nx} = \cos{(nx+nT_1)} = \cos{(nx+2\pi)} (I\quad used \quad the \quad periodicity\quad of\quad the \quad second\quad function)\\ nx+2\pi = nx+nT_1\\ T_1=\frac{2\pi}{n} $$