91Ó°ÊÓ

We have to prove that the period of sine function is $2\mathrm{Ï€}$.

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$2\mathrm{Ï€}$.

Consider the sine function as following :-

$f\left(\mathrm{θ}\right)=\sin \mathrm{θ}$.

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

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

$f(\mathrm{θ}+p)=f\left(\mathrm{θ}\right)$, for all $\mathrm{θ}$.

Now put $\mathrm{θ}=0$, then we have :-

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

We know that the sine function is 0, for $n\mathrm{π},n∈Z$.

Then we have :-

$p=n\mathrm{π},n∈Z$

We find that $p=n\mathrm{π},n∈Z$.

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

This gives us :-

$p=\mathrm{Ï€}$.

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

localid="1647152875440" $$\begin{gathered} \sin (\mathrm{θ}+p)=\sin \mathrm{θ} \\ ⇒\sin (\mathrm{θ}+\mathrm{π})=\sin \mathrm{θ} \end{gathered}$$

As we know that the value of $\sin (\mathrm{θ}+\mathrm{π})=-\sin \mathrm{θ}$, as sine is negative in third quadrant, then we have :-

$-\sin \mathrm{θ}=\sin \mathrm{θ}$.

But this is not true.

That is $\mathrm{Ï€}$is not the period of sine function.

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

Then we have :-

$p=2\mathrm{Ï€}$.

Now in $\sin (\mathrm{θ}+p)=\sin \mathrm{θ}$, put $p=2\mathrm{π}$, then we have :-

$$\begin{gathered} \sin \left(\mathrm{θ}+2\mathrm{π}\right)=\sin \mathrm{θ} \\ ⇒\sin \mathrm{θ}=\sin \mathrm{θ} \end{gathered}$$

As sine is positive in first quadrant.

This satisfies the required condition.

As we find that $p=2\mathrm{Ï€}$, is the smallest value that satisfies the periodic condition for sine function.

So we can conclude that the period of sine function is$2\mathrm{Ï€}$.