$\begin{array}{l}
{\cos ^2}\left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{1 + \cos \left( {2x + \dfrac{\pi }{4}.2} \right)}}{2}\\
= \dfrac{{1 + \cos \left( {2x + \dfrac{\pi }{2}} \right)}}{2} = \dfrac{{1 + \cos \left[ {\dfrac{\pi }{2} - \left( { - 2x} \right)} \right]}}{2}\\
= \dfrac{{1 + \sin \left( { - 2x} \right)}}{2} = \dfrac{{1 - \sin 2x}}{2} = \dfrac{{{{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x}}{2}\\
= \dfrac{{{{\left( {\sin x - \cos x} \right)}^2}}}{2}
\end{array}$
