Show that the period of$f\left(\theta \right)=\sin \theta$is$2\pi$.

We have to prove that the period of sine function is $2\pi$.

Firstly we will assume that period of sine function is $p$.

Then by finding the least value for$p$, we will show that the period of sine function is$2\pi$.

Consider the sine function as following :-

$f\left(\theta \right)=\sin \theta$.

Let us assume period of this function is $p$.

As we assume period of sine function is $p$, then it satisfies that :-

$f(\theta +p)=f\left(\theta \right)$, for all $\theta$.

Now put $\theta =0$, then we have :-

$$\begin{gathered} f(0+p)=f\left(0\right) \\ \Rightarrow \sin (0+p)=\sin 0 \\ \Rightarrow \sin \left(p\right)=\sin 0 \end{gathered}$$

We know that the sine function is 0, for $n\pi ,n\in Z$.

Then we have :-

$p=n\pi ,n\in Z$

We find that $p=n\pi ,n\in Z$.

Now to find period take smallest value of $n=1$.

This gives us :-

$p=\mathrm{\pi }$.

As we assume $p$is period of sine function, this will satisfies :-

localid="1647152875440" $$\begin{gathered} \sin (\theta +p)=\sin \theta \\ \Rightarrow \sin (\theta +\pi )=\sin \theta \end{gathered}$$

As we know that the value of $\sin (\theta +\pi )=-\sin \theta$, as sine is negative in third quadrant, then we have :-

$-\sin \theta =\sin \theta$.

But this is not true.

That is $\pi$is not the period of sine function.

Now we check for next smallest value. That is for $n=2$.

Then we have :-

$p=2\pi$.

Now in $\sin (\theta +p)=\sin \theta$, put $p=2\pi$, then we have :-

$$\begin{gathered} \sin \left(\theta +2\pi \right)=\sin \theta \\ \Rightarrow \sin \theta =\sin \theta \end{gathered}$$

As sine is positive in first quadrant.

This satisfies the required condition.

As we find that $p=2\pi$, is the smallest value that satisfies the periodic condition for sine function.

So we can conclude that the period of sine function is$2\pi$.