Prove the following trigonometric equation:
$\cos 2x\cos \dfrac{x}{2}-\cos 3x\cos \dfrac{9x}{2}=\sin 5x\sin \dfrac{5x}{2}$

Hint:To prove the above equation, we are trying to reduce the L.H.S expression to R.H.S of the given expression. We are going to use the identity of $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$ in the L.H.S expression and then simplify.

Complete step-by-step answer:
The given equation that we have to prove is:
$\cos 2x\cos \dfrac{x}{2}-\cos 3x\cos \dfrac{9x}{2}=\sin 5x\sin \dfrac{5x}{2}$
Applying the identity of $2\cos A\cos B$ on $\cos 2x\cos \dfrac{x}{2}$ we get,
$2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$
$\begin{align}
  & \cos 2x\cos \dfrac{x}{2}=\dfrac{1}{2}\left( \cos \left( 2x+\dfrac{x}{2} \right)+\cos \left( 2x-\dfrac{x}{2} \right) \right) \\
 & \Rightarrow \cos 2x\cos \dfrac{x}{2}=\dfrac{1}{2}\left( \cos \left( \dfrac{5x}{2} \right)+\cos \left( \dfrac{3x}{2} \right) \right) \\
\end{align}$
Applying the identity of $2\cos A\cos B$ on $\cos 3x\cos \dfrac{9x}{2}$ we get,
$\begin{align}
  & \cos 3x\cos \dfrac{9x}{2}=\dfrac{1}{2}\left( \cos \left( 3x+\dfrac{9x}{2} \right)+\cos \left( 3x-\dfrac{9x}{2} \right) \right) \\
 & \Rightarrow \cos 3x\cos \dfrac{9x}{2}=\dfrac{1}{2}\left( \cos \left( \dfrac{15x}{2} \right)+\cos \left( -\dfrac{3x}{2} \right) \right) \\
\end{align}$
Now, substituting these values in the L.H.S of the given expression we get,
$\begin{align}
  & \cos 2x\cos \dfrac{x}{2}-\cos 3x\cos \dfrac{9x}{2} \\
 & =\dfrac{1}{2}\left( \left( \cos \dfrac{5x}{2}+\cos \dfrac{3x}{2} \right)-\left( \cos \dfrac{15x}{2}+\cos \left( -\dfrac{3x}{2} \right) \right) \right) \\
\end{align}$
We know that $\cos \left( -x \right)=\cos x$ so applying this cosine property in the above expression,
$\begin{align}
  & \dfrac{1}{2}\left( \left( \cos \dfrac{5x}{2}+\cos \dfrac{3x}{2} \right)-\left( \cos \dfrac{15x}{2}+\cos \left( \dfrac{3x}{2} \right) \right) \right) \\
 & =\dfrac{1}{2}\left( \cos \dfrac{5x}{2}+\cos \dfrac{3x}{2}-\cos \dfrac{15x}{2}-\cos \dfrac{3x}{2} \right) \\
\end{align}$
As $\cos \dfrac{3x}{2}$ will be cancelled out in the above expression then the remaining expression will look like:
$\dfrac{1}{2}\left( \cos \dfrac{5x}{2}-\cos \dfrac{15x}{2} \right)$
Now, we are going to apply the identity of $\cos C-\cos D$ in the above expression as:
$\cos C-\cos D=-2\sin \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$
$\begin{align}
  & \dfrac{1}{2}\left( -2\sin \left( \dfrac{\dfrac{5x}{2}+\dfrac{15x}{2}}{2} \right)\sin \left( \dfrac{\dfrac{5x}{2}-\dfrac{15x}{2}}{2} \right) \right) \\
 & =\dfrac{1}{2}\left( -2\sin 5x\sin \left( -\dfrac{5x}{2} \right) \right) \\
\end{align}$
We know from the trigonometric property of sine that $\sin (-x)=-\sin x$. Applying this property in the above expression and you can also see that 2 will be cancelled out from the numerator and the denominator.
$\sin 5x\sin \dfrac{5x}{2}$
The simplification of L.H.S has come out as $\sin 5x\sin \dfrac{5x}{2}$ which is equal to R.H.S of the given expression.
Hence, we have shown that L.H.S = R.H.S of the given expression.

Note: The other way of doing this equation is:
 $\cos 2x\cos \dfrac{x}{2}-\cos 3x\cos \dfrac{9x}{2}=\sin 5x\sin \dfrac{5x}{2}$
We can rewrite the above equation as:
$\cos 2x\cos \dfrac{x}{2}=\cos 3x\cos \dfrac{9x}{2}+\sin 5x\sin \dfrac{5x}{2}$
Now, solve the R.H.S using the identities of $2\cos A\cos B$ and $2\sin A\sin B$ as follows:
$\begin{align}
  & \dfrac{1}{2}\left( \cos \left( \dfrac{15x}{2} \right)+\cos \left( -\dfrac{3x}{2} \right)+\cos \left( \dfrac{5x}{2} \right)-\cos \left( \dfrac{15x}{2} \right) \right) \\
 & =\dfrac{1}{2}\left( \cos \left( \dfrac{3x}{2} \right)+\cos \left( \dfrac{5x}{2} \right) \right) \\
\end{align}$
Now, using the identity of $\cos C+\cos D$ in the above equation we get,
$\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$
$\dfrac{1}{2}\left( 2\cos 2x\cos \dfrac{x}{2} \right)$
As 2 will be cancelled out from the numerator and the denominator so the remaining expression will look like:
$\cos 2x\cos \dfrac{x}{2}$
As we can see, that simplification of the R.H.S yields $\cos 2x\cos \dfrac{x}{2}$ which is equal to L.H.S of the given expression.
Hence, we have proved L.H.S = R.H.S of the given expression.