Question

For what integral value of n is \[3\pi \] the period of the function $\cos \left( nx \right)\sin \left( \dfrac{5x}{n} \right)$?

Answer 1

Hint: First of all we have to know about the period of the function. The period of any function $f\left( x \right)$ is T such that $f\left( x+T \right)=f\left( x \right)$. So, we will use this condition to find the required value.

Complete step-by-step answer:
We have been asked to find the integral value of n such that \[3\pi \] is the period of the function $\cos \left( nx \right)\sin \left( \dfrac{5x}{n} \right)$ .
Let, $f\left( x \right)=\cos \left( nx \right)\sin \left( \dfrac{5x}{n} \right)$
We know that if T is a period of a function $f\left( x \right)$ then $f\left( x+T \right)=f\left( x \right)$.
We have \[3\pi \] is the period of the function $f\left( x \right)$.
\[\begin{align}
  & \Rightarrow f\left( x+3\pi \right)=f\left( x \right) \\
 & \Rightarrow \cos \left( n\left( x+3\pi \right) \right)\sin \left( \dfrac{5\left( x+3\pi \right)}{n} \right)=\cos \left( nx \right)\sin \left( \dfrac{5x}{n} \right) \\
 & \Rightarrow \cos \left( nx+3n\pi \right)\sin \left( \dfrac{15\pi +5x}{n} \right)=\cos nx\sin \left( \dfrac{5x}{n} \right) \\
\end{align}\]
Since, $\cos \left( x+2n\pi \right)=\cos x\ and\ \cos \left( x+\left( 2n+1 \right)\pi \right)=-\cos x$
$\begin{align}
  & \Rightarrow {{\left( -1 \right)}^{n}}\cos \left( nx \right)\sin \left( \dfrac{15\pi }{n}+\dfrac{5x}{n} \right)=\cos nx\sin \left( \dfrac{5x}{n} \right) \\
 & \Rightarrow {{\left( -1 \right)}^{n}}\cos nx\sin \left( \dfrac{5x}{n}+\dfrac{15\pi }{n} \right)-\cos nx\sin \dfrac{5x}{n}=0 \\
\end{align}$
Taking $'\cos nx'$ as common, we get,
$\cos nx\left( {{\left( -1 \right)}^{n}}\sin \left( \dfrac{5x}{n}+\dfrac{15\pi }{n} \right)-\sin \dfrac{5x}{n} \right)=0$
Now, $\cos nx\ne 0$ because if it is equal to zero then $f\left( x \right)=0$.
$\begin{align}
  & \Rightarrow \left( {{\left( -1 \right)}^{n}}\sin \left( \dfrac{5x}{n}+\dfrac{15\pi }{n} \right)-\sin \dfrac{5x}{n} \right)=0 \\
 & \Rightarrow {{\left( -1 \right)}^{n}}\sin \left( \dfrac{5x}{n}+\dfrac{15\pi }{n} \right)=\sin \left( \dfrac{5x}{n} \right) \\
\end{align}$
By using the trigonometric identity,
$\begin{align}
  & \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B \\
 & {{\left( -1 \right)}^{n}}\left[ \sin \left( \dfrac{5x}{n} \right)\cos \dfrac{15\pi }{n}+\cos \left( \dfrac{5x}{n} \right)\sin \left( \dfrac{15\pi }{n} \right) \right]=\sin \left( \dfrac{5x}{n} \right) \\
\end{align}$
We know that the periodic function $f\left( x \right)$ is satisfied for all values of ‘x’.
Let us assume $x=0$ for our equation.
$\begin{align}
  & \Rightarrow {{\left( -1 \right)}^{n}}\left[ \sin 0\cos \dfrac{15\pi }{n}+\cos 0\sin \left( \dfrac{15\pi }{n} \right) \right]=\sin 0 \\
 & \Rightarrow {{\left( -1 \right)}^{n}}\left[ 0+\sin \left( \dfrac{15\pi }{n} \right) \right]=0 \\
 & \Rightarrow \sin \left[ \dfrac{15\pi }{n} \right]=0 \\
\end{align}$
Since, we know that $\sin \left( n\pi \right)=0$ where $n=0,1,2,3,......$
$\Rightarrow n=\pm 1,\pm 3,\pm 5,\pm 15$
Therefore, the integral values are $\pm 1,\pm 3,\pm 5\ and\ \pm 15$.

Note: Be careful while doing the simplification of the function and also take care of the sign.
Also, remember that the period of the function is the distance between repetition of any function and these types of function is known as periodic function and it satisfied the condition, $f\left( x+T \right)=f\left( x \right)$ where T is the fundamental period of the function $f\left( x \right)$.