Finding the value of $\sin {\frac{31 \pi}3}$ The task was to find out the value of $$\sin\frac{31\pi}3$$
This is a example in my book in which following steps are shown:
$$\begin{align}
\sin\frac{31\pi}3& =\sin\left(10\pi +\frac\pi3\right)\\
&=\sin\frac\pi3\\
&=\frac{\sqrt3}2
\end{align}$$
I cannot understand step 3, $=\sin\frac\pi3$
 A: The sine function is periodic with period $2\pi$. A periodic function is a function that repeats its values in regular intervals or periods. A function is said to be periodic with period $P$, where $P$ being a nonzero constant if we have
$$
f(x+P) = f(x)
$$
for all values of $x$ in the domain. Therefore
$$
\sin(n\pi+\theta)=\left\{ 
  \begin{array}{l l}
    \sin\theta & ;\quad \text{if $n$ is even},\\
\\
    -\sin\theta & ;\quad \text{if $n$ is odd}.
  \end{array} \right.
$$
A: Multiples of $2\pi$ do not matter because $\sin$ is periodic of period $2\pi$.
A: Think of the unit circle and the definition of $\sin$: A straight line drawn from the origin with the angle between the $x$-axis of $\theta$ intersects the unit circle at $(\cos \theta, \sin \theta)$. 

Say you have a point on the unit circle at $(\cos \theta, \sin \theta)$. If the line making the angle $\theta$ from the origin makes a full revolution (revolves around the origin by a measure of $2\pi$), then it will end up in the same position. Thus, $\sin n = \sin (2\pi + n) = \sin (n - 2\pi)$. 
A: You can confirm this by using the angle-sum formula:
$$\begin{align} \sin(10\pi + \pi/3) & = \sin(10\pi)\cos(\pi/3) + \cos(10 \pi)\sin(\pi/3) \\ \\ & = 0\cdot\frac 12 + 1 \cdot \sin (\pi/3) \\ \\ & = \sin(\pi/3)\\ \\ & = \sqrt 3/2\end{align}$$
A: The sine function is periodic with period $2\pi$. This means that $\sin\theta = \sin(\theta + 2\pi) \ \forall \ \theta$.  One can apply this identity five times to get rid of the $10 \pi$ term.
