Question

Find the maximum value of $1+\sin (\frac{\pi }{4}+\theta )+2\sin (\frac{\pi }{4}-\theta )$ for all real values of $\theta$

Answer 1

Let $A=1+\sin (\frac{\pi }{4}+\theta )+2\sin (\frac{\pi }{4}-\theta )$
$=1+\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta +\sqrt{2}\cos \theta -\sqrt{2}\sin \theta$
$=1+(\frac{1}{\sqrt{2}}-\sqrt{2})\sin \theta +(\frac{1}{\sqrt{2}}+\sqrt{2})$
As $-\sqrt{a^2+b^2}\le a\sin x+b\cos x\le \sqrt{a^2+b^2}$
$\therefore -5\le A\le 5$