Question

How do you find the period of $y=\sin \left( \dfrac{x}{3} \right)$ ?

Answer 1

Hint: To find the period of the above trigonometric function $y=\sin \left( \dfrac{x}{3} \right)$, we will substitute x as $\left( x+T \right)$ then the value of y when x is substituted is same as the value of y when x as $\left( x+T \right)$ has been substituted. So, “T” here is the period of the given function. To find this value of “T”, we know the period of $\sin x$ is $2\pi $ so using this we can find the period for $\sin \left( \dfrac{x}{3} \right)$.

Complete step-by-step answer:
The function given above which we have to find the period is as follows:
$y=\sin \left( \dfrac{x}{3} \right)$
To know the period of the above function, first of all, we should know the period of $\sin x$.
The period of any function is found when the below equation holds true:
$f\left( x+T \right)=f\left( x \right)$
Now, taking $f\left( x \right)=\sin x$ in the above function we get,
$\Rightarrow \sin \left( x+T \right)=\sin x$
Now, we are going to find the period of $\sin x$ and we know the period of $\sin x$ as $2\pi $ so substituting $T=2\pi $ in the above equation we get,
$\Rightarrow \sin \left( x+2\pi \right)=\sin x$
Also, from the trigonometric properties, we know that:
$\sin \left( 2\pi +\theta \right)=\sin \theta $
So, using the above relation the equation we have just written above this property holds true.
$\Rightarrow \sin \left( x+2\pi \right)=\sin x$ ……. Eq. (1)
Hence, we have shown $2\pi $ as the period of $\sin x$.
Now, to find the period of $\sin \left( \dfrac{x}{3} \right)$, we need some T so that on putting x as $\left( x+T \right)$ we get the same function as $\sin \left( \dfrac{x}{3} \right)$. We are writing mathematically what we have just described we get,
$\Rightarrow \sin \left( \dfrac{x+T}{3} \right)=\sin \left( \dfrac{x}{3} \right)$
Rearranging the expression written in the bracket in the L.H.S of the above equation we get,
$\Rightarrow \sin \left( \dfrac{x}{3}+\dfrac{T}{3} \right)=\sin \left( \dfrac{x}{3} \right)$
Now, the above equation holds true when $\dfrac{T}{3}$ becomes $2\pi $. This we can say from eq. (1) so equating $\dfrac{T}{3}$ to $2\pi $ we get,
$\Rightarrow \dfrac{T}{3}=2\pi $
Cross multiplying the above equation we get,
$\begin{align}
  & \Rightarrow T=3\left( 2\pi \right) \\
 & \Rightarrow T=6\pi \\
\end{align}$
Hence, we got the period of $\sin \left( \dfrac{x}{3} \right)$ as $6\pi $.

Note: The similar problem which can be possible is that you might ask to find the period for $\cos \left( \dfrac{x}{3} \right)$ which is the same as that of $\sin \left( \dfrac{x}{3} \right)$ because for $\cos \left( \dfrac{x}{3} \right)$, we need the period for $\cos x$ and the period of $\cos x$ is same as that of $\sin x$ i.e. $2\pi $ so the period for $\cos \left( \dfrac{x}{3} \right)$ is same as that of $\sin \left( \dfrac{x}{3} \right)$ which is equal to $6\pi $.