Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors, then the greatest value of $\sqrt{3}\left|\begin{array}{c}\overrightarrow{a}+\overrightarrow{b}\end{array}\right|+\left|\begin{array}{c}\overrightarrow{a}-\overrightarrow{b}\end{array}\right|$ is

Answer 1

$\sqrt{3}\left|\begin{array}{c}\overrightarrow{a}+\overrightarrow{b}\end{array}\right|+\left|\begin{array}{c}\overrightarrow{a}-\overrightarrow{b}\end{array}\right|$
$=\sqrt{3}\left(\sqrt{2+2\cos \theta }\right)+\sqrt{2-2\cos \theta }$
$=\sqrt{6}\left(\sqrt{1+\cos \theta }\right)+\sqrt{2}\left(\sqrt{1-\cos \theta }\right)$
$=\sqrt{6}\left(\sqrt{1+2{\cos }^2\frac{\theta }{2}-1}\right)+\sqrt{2}\left(\sqrt{1-1+2{\sin }^2\frac{\theta }{2}}\right)$
$=\sqrt{6}\left(\sqrt{2{\cos }^2\frac{\theta }{2}}\right)+\sqrt{2}\left(\sqrt{2{\sin }^2\frac{\theta }{2}}\right)$
$\left\{\text{greatest value of trigonometric function}\right\}a\cos \theta +b\sin \theta \le \sqrt{a^2+b^2}$
$=2\sqrt{3}\left|\begin{array}{c}\cos \frac{\theta }{2}\end{array}\right|+2\left|\begin{array}{c}\sin \frac{\theta }{2}\end{array}\right|\le \sqrt{(2\sqrt{3}{)}^2+(2{)}^2}=4$