Question

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

Answer 1

Answer: 4

$\Rightarrow 1+\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta +2\left(\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta \right)$
$\Rightarrow 1+\left(\frac{1}{\sqrt{2}}+\sqrt{2}\right)\sin \theta +\left(\frac{1}{\sqrt{2}}+\sqrt{2}\right)\cos \theta$
$\Rightarrow$ Maximum value is
$\sqrt{{\left(\frac{3}{\sqrt{2}}\right)}^2+{\left(\frac{3}{\sqrt{2}}\right)}^2}+1=\sqrt{\frac{9}{2}+\frac{9}{2}}+1=4$