Proof that $ \frac {2\pi} {k} $ is the period of $ y=\sin(kx) $ How can you prove that $ \frac {2\pi} {k} $ is the period of the function $ y=\sin(kx) $?
 A: Recall sine is a periodic function. For the function $y = \sin b(x)$, $b$ represents frequency, or rather, the number of cycles in the domain $0 \leq x \leq 2\pi$.
The function is periodic, meaning that the process repeats every $2\pi$ radians. The period or wavelength of the function is given by $\frac{2\pi}{b}$. (Because $b$ cycles of a wave with a wavelength $\frac{2\pi}{b}$ would cover $2\pi$ radians.)
You can also try factoring $bx+2\pi n$ to demonstrate this.
For all $n \in Z$:
$$\sin x = \sin(x+\color{blue}{2\pi n})$$
$$\sin (\color{purple}{b}x) = \sin (\color{purple}{b}x+\color{blue}{2\pi n}) = \sin \color{purple}{b}\bigg(x+\color{blue}{\frac{2\pi}{b}n}\bigg)$$
$$\implies \sin \color{purple}{b}(x) = \sin \color{purple}{b}\bigg(x+\color{blue}{\frac{2\pi}{b}n}\bigg)$$
This way, it becomes clear what the period of the function is.
A: HINT
Recall that by definition $\sin x$ has period $2\pi$ that is $$\sin (x)=\sin (x+2\pi)$$
A: HINT
A function is called periodic with the period $P$ if it is true for all $x$ that  $f(x+P)=f(x)$.
Therefore consider your function $\sin(kx)$ and set $x=x+\frac{2\pi}{k}$. So you will get
$$\sin(kx)=\sin\left(k\left[x+\frac{2\pi}{k}\right]\right)=\sin(kx+2\pi)$$
and now recall the basic periodic properties of the trigonometric sine function.
A: The period of the function $y$ is the smallest positive number $T$ such that $y(x+T)=y(x)\;\forall x$.
here the smallest positive number $T$ such that $\sin\bigl(k(x+T)\bigl)=\sin(kx)$ for all $x$, which by substitution is obviously the same problem as the smallest positive number $T'=kT$ such that $\sin(kx+T')=\sin kx$ for all $kx$. So
$$T'=2\pi,\quad\text{whence}\quad kT=2\pi.$$
