# What is the frequency of f(theta)= sin 18 t - cos 9 t ?

Jul 16, 2018

The frequency is $f = \frac{9}{2 \pi} H z$

#### Explanation:

First determine the period $T$

The period $T$ of a periodic function $f \left(x\right)$ is defined by

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

Here,

$f \left(t\right) = \sin \left(18 t\right) - \cos \left(9 t\right)$............................$\left(1\right)$

Therefore,

$f \left(t + T\right) = \sin \left(18 \left(t + T\right)\right) - \cos \left(9 \left(t + T\right)\right)$

$= \sin \left(18 t + 18 T\right) - \cos \left(9 t + 9 T\right)$

$= \sin 18 t \cos 18 T + \cos 18 T \sin 18 t - \cos 9 t \cos 9 T + \sin 9 t \sin 9 T$

Comparing $f \left(t\right)$ and $f \left(t + T\right)$

$\left\{\begin{matrix}\cos 18 T = 1 \\ \sin 18 T = 0 \\ \cos 9 T = 1 \\ \sin 9 T = 0\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}18 T = 2 \pi \\ 9 T = 2 \pi\end{matrix}\right.$

$\implies$, ${T}_{1} = \frac{\pi}{9}$ and ${T}_{2} = \frac{2}{9} \pi$

The $L C M$ of ${T}_{1}$ and ${T}_{2}$ is $T = \frac{2}{9} \pi$

Therefore,

The frequency is

$f = \frac{1}{T} = \frac{9}{2 \pi} H z$

graph{sin(18x)-cos(9x) [-2.32, 4.608, -1.762, 1.703]}