# solve the following If cos 2θ = 0,

Question:

If $\cos 2 \theta=0$, then $\left|\begin{array}{ccc}0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right|^{2}=$___________

Solution:

Given: $\cos 2 \theta=0$

$\cos 2 \theta=0$

$\Rightarrow \cos ^{2} \theta-\sin ^{2} \theta=0$

$\Rightarrow \cos ^{2} \theta=\sin ^{2} \theta$

$\Rightarrow \theta=\pm \frac{\pi}{4} \quad \ldots(1)$

Now,

$\left|\begin{array}{ccc}0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right|^{2}$

Expanding along $R_{1}$

$=\left[-\cos \theta\left(\cos ^{2} \theta-0\right)+\sin \theta\left(0-\sin ^{2} \theta\right)\right]^{2}$

$=\left[-\cos \theta\left(\cos ^{2} \theta\right)+\sin \theta\left(-\sin ^{2} \theta\right)\right]^{2}$

$=\left[-\cos ^{3} \theta-\sin ^{3} \theta\right]^{2}$

$=\left[-\left(\cos ^{3} \theta+\sin ^{3} \theta\right)\right]^{2}$

$=\left(\cos ^{3} \theta+\sin ^{3} \theta\right)^{2}$

$=\left((\cos \theta+\sin \theta)\left(\cos ^{2} \theta+\sin ^{2} \theta-\cos \theta \sin \theta\right)\right)^{2}$

$=((\cos \theta+\sin \theta)(1-\cos \theta \sin \theta))^{2}$

For $\theta=\frac{\pi}{4}$

$\left|\begin{array}{ccc}0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right|^{2}=\left(\left(\cos \frac{\pi}{4}+\sin \frac{\pi}{4}\right)\left(1-\cos \frac{\pi}{4} \sin \frac{\pi}{4}\right)\right)^{2}$

$=\left(\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\right)\right)^{2}$

$=\left(\left(\frac{2}{\sqrt{2}}\right)\left(1-\frac{1}{2}\right)\right)^{2}$

$=\left(\left(\frac{2}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\right)^{2}$

$=\left(\frac{1}{\sqrt{2}}\right)^{2}$

$=\frac{1}{2}$

For $\theta=-\frac{\pi}{4}$

$\left|\begin{array}{ccc}0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right|^{2}=\left(\left(\cos \left(-\frac{\pi}{4}\right)+\sin \left(-\frac{\pi}{4}\right)\right)\left(1-\cos \left(-\frac{\pi}{4}\right) \sin \left(-\frac{\pi}{4}\right)\right)\right)^{2}$

$=\left(\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{2}} \times \frac{-1}{\sqrt{2}}\right)\right)^{2}$

$=\left((0)\left(1+\frac{1}{2}\right)\right)^{2}$

$=0$

Hence, if $\cos 2 \theta=0$, then $\left|\begin{array}{ccc}0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right|^{2}=\underline{0 \text { or } \frac{1}{2}}$