Maximum value of $\sin \mathrm{xcos}(\mathrm{\pi }/4-\mathrm{x}) \mathrm{is}$

$$\begin{gathered} \sin \mathrm{xcos}\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)=\sin \mathrm{x}\left[\cos \left(\frac{\mathrm{\pi }}{4}\right)\cos \mathrm{x}+\sin \left(\frac{\mathrm{\pi }}{4}\right)\sin \mathrm{x}\right]=\frac{1}{\sqrt{2}}\left[\sin \mathrm{xcos}\mathrm{x}+{\sin }^2\mathrm{x}\right] \\ =\frac{1}{\sqrt{2}}\left[\frac{1}{2}\sin 2\mathrm{x}+\frac{1-\cos 2\mathrm{x}}{2}\right]=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}[\sin 2\mathrm{x}-\cos 2\mathrm{x}] \end{gathered}$$
Maximum value $=\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}\left(\sqrt{2}\right)=\frac{\sqrt{2}+1}{2\sqrt{2}}$