Question

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

Answer 1

`2pi`

Explanation:

Period of sin 2t --> `(2pi)/2 = pi`
Period of cos 5t -->`(2pi)/5`
Period of f(t) --> least common multiple of `pi and (2pi)/5.`

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

Period of f(t) is `(2pi)`

Answer 2

The frequency is `=1/(2pi)`

Explanation:

The frequency is `f=1/T`

The period is `=T`

A function `f(theta)` is T-periodic iif

`f(theta)=(theta+T)`

Therefore,

`sin(2t)-cos(5t)=sin2(t+T)-cos5(t+T)`

Therefore,

`{(sin (2t)=sin2(t+T)),(cos(5t)=cos5(t+T)):}`

`<=>`, `{(sin2t=sin(2t+2T)),(cos5t=cos(5t+5T)):}`

`<=>`, `{(sin2t=sin2tcos2T+cos2tsin2T),(cos5t=cos5tcos5T-sin5tsin5T):}`

`<=>`, `{(cos2T=1),(cos5T=1):}`

`<=>`, `{(2T=2pi=4pi),(5T=2pi=4pi=6pi=8pi=10pi):}`

`<=>`, `{(T=4/2pi=2pi),(T=10/5pi=2pi):}`

The period is `=2pi`

The frequency is

`f=1/(2pi)`

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