Prove the following expression, ${{\cos }^{2}}48{}^\circ -{{\sin }^{2}}12{}^\circ =\dfrac{\sqrt{5}+1}{8}$.

Hint: To prove the relation, we should know that $\sin \theta =\cos \left( 90{}^\circ -\theta \right)$. And we should also know a few addition formulas of trigonometry like, $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and $\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$. We should also know, ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$. By using these relations, we can prove the required result.

Complete Step-by-Step solution:
To prove this relation, we will first consider the left hand side or the LHS of the desired expression, that is, ${{\cos }^{2}}48{}^\circ -{{\sin }^{2}}12{}^\circ $. We know that, $\sin \theta =\cos \left( 90{}^\circ -\theta \right)$, so by applying that, we get LHS as,
$\begin{align}
  & ={{\cos }^{2}}48{}^\circ -{{\cos }^{2}}\left( 90-12{}^\circ \right) \\
 & ={{\cos }^{2}}48{}^\circ -{{\cos }^{2}}78{}^\circ \\
\end{align}$
Now, we know that ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$, so by applying that in the above expression, we get the LHS as,
$=\left( \cos 48{}^\circ +\cos 78{}^\circ \right)\left( \cos 48{}^\circ -\cos 78{}^\circ \right)$
Now, we know that $\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$ and $\cos A-\cos B=-2\sin \left( \dfrac{A+B}{2} \right)\sin \left( \dfrac{A-B}{2} \right)$, so, by applying that, we get the LHS as,
\[\begin{align}
  & =\left[ 2\cos \left( \dfrac{48{}^\circ +78{}^\circ }{2} \right)\cos \left( \dfrac{48{}^\circ -78}{2}{}^\circ \right) \right]\times \left[ -2\sin \left( \dfrac{48{}^\circ +78{}^\circ }{2} \right)\sin \left( \dfrac{48{}^\circ -78{}^\circ }{2} \right) \right] \\
 & =\left[ 2\left( \cos 63{}^\circ \right)\left( \cos 15{}^\circ \right) \right]\times \left[ -2\left( \sin 63{}^\circ \right)\left( \sin \left( -15{}^\circ \right) \right) \right] \\
\end{align}\]
By rearranging the terms, we get,
\[=\left[ 2\left( \sin 63{}^\circ \right)\left( \cos 63{}^\circ \right) \right]\times \left[ -2\sin \left( -15{}^\circ \right)\cos \left( 15{}^\circ \right) \right]\]
We know that $\sin \left( -\theta \right)=-\sin \theta $ and $2\sin \theta \cos \theta =\sin 2\theta $, so by applying that, we get LHS as,
$\begin{align}
  & =\left[ 2\left( \sin 63{}^\circ \right)\left( \cos 63{}^\circ \right) \right]\times \left[ 2\left( \sin 15{}^\circ \right)\left( \cos 15{}^\circ \right) \right] \\
 & =\sin 2\left( 63{}^\circ \right)\times \sin 2\left( 15{}^\circ \right) \\
 & =\sin \left( 126{}^\circ \right)\times \sin \left( 30{}^\circ \right) \\
\end{align}$
We know that $\sin 30{}^\circ =\dfrac{1}{2}$ and that $\sin \left( 90+\theta \right)=\cos \theta $. So, by applying these and rearranging the terms, we get the LHS as,
$\begin{align}
  & =\sin 30{}^\circ \sin \left( 90{}^\circ +36{}^\circ \right) \\
 & =\dfrac{1}{2}\cos 36{}^\circ \\
\end{align}$
We should know that $\cos 36{}^\circ =\dfrac{\sqrt{5}+1}{4}$. So, substituting the same in the above we get the LHS as,
$\begin{align}
  & =\dfrac{1}{2}.\dfrac{\sqrt{5}+1}{4} \\
 & =\dfrac{\sqrt{5}+1}{8} \\
\end{align}$
Which is equal to what is given in the right-hand side of the expression to be proved in the question.
Hence, we have thus proved the expression given in the question, that is, ${{\cos }^{2}}48{}^\circ -{{\sin }^{2}}12{}^\circ =\dfrac{\sqrt{5}+1}{8}$.

Note: We can also find the value of $\cos 36{}^\circ $ by assuming its value as $18{}^\circ =A$ and $90{}^\circ =5A$. This is nothing but $90{}^\circ =2A+3A$ or we can also say it as, $2A=90{}^\circ -3A$. Now we will take sin to both the sides and will form a quadratic equation. By doing so, we will get, $\sin 18{}^\circ =\dfrac{-1+\sqrt{5}}{4}$. We know that $\cos 2\theta =1-2{{\sin }^{2}}\theta $. So, we will get, $\cos 36{}^\circ =\dfrac{\sqrt{5}+1}{4}$.