Four simple harmonic vibrations are: ${{y}_{1}}=8\cos \omega t$, ${{y}_{2}}=4\cos \left( \omega t+\dfrac{\pi }{2} \right)$, ${{y}_{3}}=2\cos \left( \omega t+\pi \right)$, ${{y}_{4}}=\cos \left( \omega t+\dfrac{3\pi }{2} \right)$, magnitude and direction of resultant is:
A.$\sqrt{45}\text{ and }{{\tan }^{-1}}\left( \dfrac{1}{2} \right)$
B.$\sqrt{45}\text{ and }{{\tan }^{-1}}\left( \dfrac{1}{3} \right)$
C.$\sqrt{75}\text{ and }{{\tan }^{-1}}\left( 2 \right)$
D.$\sqrt{75}\text{ and }{{\tan }^{-1}}\left( \dfrac{1}{3} \right)$

Hint: To solve this problem, first we need to simplify the given equations in the question into a simpler form to ease out the question by using trigonometric identities. After that we can proceed to calculate the resultant and direction of the resultant wave with the help of superposition principle.

Complete answer:
Let us first start by simplifying the equations by using trigonometric identities that are known to us. There is no need to simplify ${{y}_{1}}$, because it is already in the simplest form. Let us simplify the others to make our work easier:
For the simple harmonic vibration ${{y}_{2}}$, its simplest form will be:
$\begin{align}
  & {{y}_{2}}=4\cos \left( \omega t+\dfrac{\pi }{2} \right) \\
 & \Rightarrow -4\sin \left( \omega t \right) \\
\end{align}$
For the simple harmonic vibration ${{y}_{3}}$ its simplest form will be as follows:
$\begin{align}
  & {{y}_{3}}=2\cos \left( \omega t+\pi \right) \\
 & \Rightarrow {{y}_{3}}=-2\cos \left( \omega t \right) \\
\end{align}$
For the simple harmonic vibration ${{y}_{4}}$ its simplest form will be as follows:
$\begin{align}
  & {{y}_{4}}=\cos \left( \omega t+\dfrac{3\pi }{2} \right) \\
 & \Rightarrow {{y}_{4}}=\sin \left( \omega t \right) \\
\end{align}$
The resultant of the given simple harmonic vibrations will be the summation of all the simple harmonic vibrations according to the superposition theorem and will be as follows:
\[\begin{align}
  & y={{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}} \\
 & \Rightarrow y=8\cos \omega t-4\sin \omega t-2\cos \omega t+\sin \omega t \\
 & \Rightarrow y=6\cos \omega t-3\sin \omega t \\
\end{align}\]
The magnitude of the resultant simple harmonic vibrations will be:
\[\begin{align}
  & \left| y \right|=\sqrt{{{6}^{2}}+{{3}^{2}}} \\
 & \Rightarrow \left| y \right|=\sqrt{36+9} \\
 & \therefore \left| y \right|=\sqrt{45} \\
\end{align}\]
The direction of the resultant of the four simple harmonic vibrations will be as follows:
$\begin{align}
  & \tan \phi =\dfrac{3}{6} \\
 & \Rightarrow \tan \phi =\dfrac{1}{2} \\
 & \therefore \phi ={{\tan }^{-1}}\left( \dfrac{1}{2} \right) \\
\end{align}$

Thus, the correct option is $A$.

Note:
There might be a confusion on the positive signs and the negative signs while converting the sine into cosine and vice versa, we need to remember that all trigonometric functions are positive in first quadrant, only sine is positive in second quadrant, only tan in positive in third quadrant and only cosine is positive in the fourth quadrant.