# What is the frequency of f(theta)= sin 2 t - cos 5 t ?

Apr 28, 2018

$2 \pi$

#### Explanation:

Period of sin 2t --> $\frac{2 \pi}{2} = \pi$
Period of cos 5t -->$\frac{2 \pi}{5}$
Period of f(t) --> least common multiple of $\pi \mathmr{and} \frac{2 \pi}{5.}$

pi .............x 2 ... --> 2pi
(2pi)/5 ....x 5 ......--> 2pi

Period of f(t) is $\left(2 \pi\right)$

Apr 28, 2018

The frequency is $= \frac{1}{2 \pi}$

#### Explanation:

The frequency is $f = \frac{1}{T}$

The period is $= T$

A function $f \left(\theta\right)$ is T-periodic iif

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

Therefore,

$\sin \left(2 t\right) - \cos \left(5 t\right) = \sin 2 \left(t + T\right) - \cos 5 \left(t + T\right)$

Therefore,

$\left\{\begin{matrix}\sin \left(2 t\right) = \sin 2 \left(t + T\right) \\ \cos \left(5 t\right) = \cos 5 \left(t + T\right)\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}\sin 2 t = \sin \left(2 t + 2 T\right) \\ \cos 5 t = \cos \left(5 t + 5 T\right)\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}\sin 2 t = \sin 2 t \cos 2 T + \cos 2 t \sin 2 T \\ \cos 5 t = \cos 5 t \cos 5 T - \sin 5 t \sin 5 T\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}\cos 2 T = 1 \\ \cos 5 T = 1\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}2 T = 2 \pi = 4 \pi \\ 5 T = 2 \pi = 4 \pi = 6 \pi = 8 \pi = 10 \pi\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}T = \frac{4}{2} \pi = 2 \pi \\ T = \frac{10}{5} \pi = 2 \pi\end{matrix}\right.$

The period is $= 2 \pi$

The frequency is

$f = \frac{1}{2 \pi}$

graph{sin(2x)-cos(5x) [-3.75, 18.75, -7.045, 4.205]}