# What is the period of f(t)=sin( t /13 )+ cos( (13t)/24 ) ?

##### 2 Answers
Jun 11, 2018

The period is $= 4056 \pi$

#### Explanation:

The period $T$ of a periodic functon is such that

$f \left(t\right) = f \left(t + T\right)$

Here,

$f \left(t\right) = \sin \left(\frac{1}{13} t\right) + \cos \left(\frac{13}{24} t\right)$

Therefore,

$f \left(t + T\right) = \sin \left(\frac{1}{13} \left(t + T\right)\right) + \cos \left(\frac{13}{24} \left(t + T\right)\right)$

$= \sin \left(\frac{1}{13} t + \frac{1}{13} T\right) + \cos \left(\frac{13}{24} t + \frac{13}{24} T\right)$

$= \sin \left(\frac{1}{13} t\right) \cos \left(\frac{1}{13} T\right) + \cos \left(\frac{1}{13} t\right) \sin \left(\frac{1}{13} T\right) + \cos \left(\frac{13}{24} t\right) \cos \left(\frac{13}{24} T\right) - \sin \left(\frac{13}{24} t\right) \sin \left(\frac{13}{24} T\right)$

As,

$f \left(t\right) = f \left(t + T\right)$

$\left\{\begin{matrix}\cos \left(\frac{1}{13} T\right) = 1 \\ \sin \left(\frac{1}{13} T\right) = 0 \\ \cos \left(\frac{13}{24} T\right) = 1 \\ \sin \left(\frac{13}{24} T\right) = 0\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}\frac{1}{13} T = 2 \pi \\ \frac{13}{24} T = 2 \pi\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}T = 26 \pi = 338 \pi \\ T = \frac{48}{13} \pi = 48 \pi\end{matrix}\right.$

$\iff$, $T = 4056 \pi$

Jun 11, 2018

$624 \pi$

#### Explanation:

Period of $\sin \left(\frac{t}{13}\right)$ --> $13 \left(2 \pi\right) = 26 \pi$
Period of $\cos \left(\frac{13 t}{24}\right)$ --> $\frac{\left(24\right) \left(2 \pi\right)}{13} = \frac{48 \pi}{13}$
Period of f(t) --> least common multiple of $26 \pi$ and $\frac{48 \pi}{13}$

$26 \pi$ .... x (24).............--> .$624 \pi$
$\frac{48 \pi}{13}$ .....x (13)(13)...--> $624 \pi$...-->
Period of f(t) --> $624 \pi$