Question

The period of $\sin 4x+\cos 4x$ is,
A. $\dfrac{{{\pi }^{4}}}{2}$
B. $\dfrac{{{\pi }^{2}}}{2}$
C. $\dfrac{\pi }{4}$
D. $\dfrac{\pi }{2}$

Answer 1

Hint: Consider $\sin 4x+\cos 4x$ as $f\left( x \right)$ and $\sin 4x$ as $g\left( x \right)$ and $\cos 4x$ as $h\left( x \right)$. So, $f\left( x \right)$ can be represented as $g\left( x \right)+h\left( x \right)$. Now we have to find the periods independently using the formula. If the function is \[\sin kx\], then its period will be $\dfrac{2\pi }{k}$. Similarly, if the function is $\cos kx$, then its period will be $\dfrac{2\pi }{k}$. After finding out the periods of both separately, take their LCM to get the answer.

Complete step-by-step solution -
We have been given a function, $\sin 4x+\cos 4x$ and we have to find its period. Let us consider $\sin 4x+\cos 4x$ as $f\left( x \right)$ which is the sum of two functions $g\left( x \right)$ and $h\left( x \right)$ which represent $\sin 4x$ and \[\cos 4x\] respectively. So, it can be written as,
$f\left( x \right)=g\left( x \right)+h\left( x \right)$
Now, we will find the period of $g\left( x \right)$ which is equal to $\sin 4x$. Using the formula, if the function is in the form of $\sin \left( kx \right)$, where $k$ is the constant, then its period will be $\dfrac{2\pi }{k}$. So, the period of $g\left( x \right)$ will be $\dfrac{2\pi }{4}$ which is $\dfrac{\pi }{2}$. Now we will check the periodicity by using,
$\begin{align}
  & g\left( x+\dfrac{\pi }{2} \right) \\
 & =\sin 4\left( x+\dfrac{\pi }{2} \right) \\
\end{align}$
By simplifying, we get,
$=\sin \left( 4x+2\pi \right)$
By using the identity, $\sin \left( 2\pi +\theta \right)=\sin \theta $, we get,
$\sin \left( 4x+2\pi \right)=\sin 4x$, which is $g\left( x \right)$. So,
$g\left( x+\dfrac{\pi }{2} \right)=g\left( x \right)$
Hence its period is $\dfrac{\pi }{2}$.
Now we will find the periodicity of $h\left( x \right)$ which is equal to $\cos 4x$. Using the formula, if the function is in the form of $\cos kx$, then its period will be $\dfrac{2\pi }{k}$. So, its period will be $\dfrac{2\pi }{4}$ which is $\dfrac{\pi }{4}$. Now we will check its periodicity. So,
$\begin{align}
  & h\left( x+\dfrac{\pi }{2} \right)=\cos 4\left( x+\dfrac{\pi }{2} \right) \\
 & =\cos \left( 4x+2\pi \right) \\
\end{align}$
Now, by using the identity, $\cos \left( 2\pi +\theta \right)=\cos \theta $, we get,
$\cos \left( 2\pi +4x \right)=\cos 4x$, which is $h\left( x \right)$. So,
$h\left( x+\dfrac{\pi }{2} \right)=h\left( x \right)$
Hence its period is $\dfrac{\pi }{2}$.
The periods for both the functions, $g\left( x \right)$ and $h\left( x \right)$ is $\dfrac{\pi }{2}$, so the LCM of both the functions is $\dfrac{\pi }{2}$.
Therefore, the correct option is D.

Note: You should be careful while finding out the periods of the functions, $g\left( x \right)$ and $h\left( x \right)$ separately as these are the places where the students tend to make mistakes. The calculations should also be done properly to avoid any mistakes.