Question

If $A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$ and $|A+A^T|=1$, then find the value(s) of $\theta$, where $\theta \in [\frac{3\pi }{2},2\pi ]$

Answer 1

$A=\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
$A^T=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]$

$A+A^T=\left[\begin{array}{cc}2\cos \theta & 0\\ 0& 2\cos \theta \end{array}\right]$
$\Rightarrow |A+A^T|=4{\cos }^2\theta$

Given, $|A+A^T|=1$
$\Rightarrow 4{\cos }^2\theta =1$
$\Rightarrow \cos \theta =\pm \frac{1}{2}$
$\Rightarrow \theta =2n\pi \pm \frac{\pi }{3}$ or $\theta =2n\pi \pm \frac{2\pi }{3}$
$\Rightarrow \theta =\frac{5\pi }{3}$ as $\theta \in [\frac{3\pi }{2},2\pi ]$