# How do I find the period of the graph, f(x)=sqrt(sin^2(x/4), without graphing?

## f(x)=sqrt(sin^2(x/4)

Jun 22, 2018

The period of $f$ is $\rho = 4 n \pi$ for any integer $n$.

#### Explanation:

Let $\rho$ be the general period of $f$. The principal period, ${\rho}_{0}$ is defined the smallest positive period of $f$.

$\rho = n {\rho}_{0}$, $n \in \mathbb{Z}$.

We ask what is the period of:

$f \left(x\right) = \sqrt{{\sin}^{2} \left(\frac{x}{4}\right)} = | \sin \left(\frac{x}{4}\right) |$

Basically, this means solving the equality below for ${\rho}_{0}$:

$f \left(x + {\rho}_{0}\right) = f \left(x\right)$

$| \sin \left(\frac{x + {\rho}_{0}}{4}\right) | = | \sin \left(\frac{x}{4}\right) |$

If $| a | = | b |$, then $a = b \mathmr{and} a = - b$.

1. $\sin \left(\frac{x + {\rho}_{0}}{4}\right) = \sin \left(\frac{x}{4}\right)$

$\sin \left(\frac{x}{4} + {\rho}_{0} / 4\right) = \sin \left(\frac{x}{4}\right)$

$\sin \left(\frac{x}{4}\right) \cos \left({\rho}_{0} / 4\right) + \cos \left(\frac{x}{4}\right) \sin \left({\rho}_{0} / 4\right) = \sin \left(\frac{x}{4}\right)$

$\implies \left\{\begin{matrix}\cos \left({\rho}_{0} / 4\right) = 1 \\ \sin \left({\rho}_{0} / 4\right) = 0\end{matrix}\right.$

With lowest positive value ${\rho}_{0} = 8 \pi$.

2. $\sin \left(\frac{x + {\rho}_{0}}{4}\right) = - \sin \left(\frac{x}{4}\right)$

$\sin \left(\frac{x}{4}\right) \cos \left({\rho}_{0} / 4\right) + \cos \left(\frac{x}{4}\right) \sin \left({\rho}_{0} / 4\right) = - \sin \left(\frac{x}{4}\right)$

$\implies \left\{\begin{matrix}\cos \left({\rho}_{0} / 4\right) = - 1 \\ \sin \left({\rho}_{0} / 4\right) = 0\end{matrix}\right.$

This time, the lowest positive value is ${\rho}_{0} = 4 \pi$.

As such, the principal period is actually

${\rho}_{0} = 4 \pi$

With the set of periods

$\mathcal{P} = \left\{4 n \pi | n \in \mathbb{Z}\right\}$