Prove the following trigonometric equation:
$\dfrac{1}{\sqrt{2}}\cos \left( \dfrac{\pi }{4}+x \right)=\dfrac{1}{2}\left( \cos (x)-\sin (x) \right)$

Hint:In this case, the cos function is to be evaluated for a sum of angles. Therefore, we can expand the Left Hand Side (LHS) of the equation using the formula for $\cos \left( A+B \right)$, where A and B are $\left( \dfrac{\pi }{4} \right)$ and $x$ respectively. As we know the value of $\cos \left( \dfrac{\pi }{4} \right)$, we can use its value and further simplify the equation to arrive at the required answer.

Complete step-by-step answer:
We have to prove that
 $\dfrac{1}{\sqrt{2}}\cos \left( \dfrac{\pi }{4}+x \right)=\dfrac{1}{2}\left( \cos (x)-\sin (x) \right)$.
We will use the following formula which gives the cosine of addition of two angles:
$\cos (a+b)=\cos (a)\cos (b)-\sin (a)\sin (b)............(1.1)$.
Now, we can simplify the left hand side of the given equation using equation (1.1) as
\[\begin{align}
  & LHS=\dfrac{1}{\sqrt{2}}\cos \left( \dfrac{\pi }{4}+x \right) \\
 & =\dfrac{1}{\sqrt{2}}\left\{ \cos \left( \dfrac{\pi }{4} \right)\cos \left( x \right)-\sin \left( \dfrac{\pi }{4} \right)\sin \left( x \right) \right\}\text{ }\left( \text{using equation (1}\text{.1)} \right)..............(1.3) \\
\end{align}\]
Now, we know that that $\pi $ corresponds to an angle of ${{180}^{\circ }}$. In the question we are given a term $\dfrac{\pi }{4}$. So, the angle corresponding to $\dfrac{\pi }{4}$ should be equal to $\dfrac{{{180}^{\circ }}}{4}={{45}^{\circ }}$.
Also, from theory of trigonometry, cosine of ${{45}^{\circ }}$ and sine of ${{45}^{\circ }}$ is equal to $\dfrac{1}{\sqrt{2}}$, i.e. $\cos {{45}^{\circ }}=\sin {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$…………….(1.2).

Therefore, using the value of $\cos ({{45}^{\circ }})$ and $\sin ({{45}^{\circ }})$ in equation (1.3) to obtain
\[\begin{align}
  & LHS=\dfrac{1}{\sqrt{2}}\left\{ \cos \left( \dfrac{\pi }{4} \right)\cos \left( x \right)-\sin \left( \dfrac{\pi }{4} \right)\sin \left( x \right) \right\}\text{ } \\
 & =\dfrac{1}{\sqrt{2}}\left( \dfrac{1}{\sqrt{2}}\times \cos x-\dfrac{1}{\sqrt{2}}\sin x \right) \\
 & =\dfrac{1}{\sqrt{2}}\times \dfrac{1}{\sqrt{2}}\left( \cos x-\sin x \right) \\
 & =\dfrac{1}{2}\left( \cos x-\sin x \right) \\
\end{align}\]
We find that the last expression is the same as given in the right hand side of the question. Thus, we have proved the equation
$\dfrac{1}{\sqrt{2}}\cos \left( \dfrac{\pi }{4}+x \right)=\dfrac{1}{2}\left( \cos (x)-\sin (x) \right)$.

Note: We should note that in the formula for cosine of the sum of two angles in equation (1.1), there is a minus sign before the second term whereas in the expansion of the sine of sum of two angles, both the terms have positive sign. Thus, one should be careful while writing the signs of the terms in the expansion formulas.Students should remember important angles in trigonometric and formulas for solving these types of problems.