The maximum value of $1+\sin \left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 4}}+\mathrm{\theta }\right)+2\cos \left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 4}}-\mathrm{\theta }\right)$  for real values of $\theta$, is 

We have,

$\begin{array}{rlrl} & \Rightarrow & \mathrm{y}& =1+\sin (\frac{\mathrm{\pi }}{4}+\mathrm{\theta })+2\cos (\frac{\mathrm{\pi }}{4}-\mathrm{\theta })\\ & \Rightarrow & \mathrm{y}& =1+\frac{1}{\sqrt{2}}(\sin \mathrm{\theta }+\cos \mathrm{\theta })+\sqrt{2}(\cos \mathrm{\theta }+\sin \mathrm{\theta })\\ & \Rightarrow & \mathrm{y}& =1+(\sqrt{2}+\frac{1}{\sqrt{2}})\mathrm{sin\theta }+(\sqrt{2}+\frac{1}{\sqrt{2}})\mathrm{cos\theta }\\ & \Rightarrow & \mathrm{y}& =1+(\sqrt{2}+\frac{1}{\sqrt{2}})(\mathrm{cos\theta }+\mathrm{sin\theta }) (\because \cos \mathrm{\theta }+\sin \mathrm{\theta }\le \sqrt{2}]\\ & \Rightarrow & \mathrm{y}\le 1+(\sqrt{2}+\frac{1}{\sqrt{2}})& \sqrt{2}\Rightarrow \mathrm{y}\le 4\end{array}$

Hence, the maximum value is 4.