<s>[INST] If $\cos2\theta=0$, then $\Delta^2=$? I'll state the question from my textbook here:

If $\cos2\theta=0$, then $\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2=$?

This is how I solved the problem:
$\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2$
$= (\cos^3 \theta + \sin^3 \theta)^2$
$= (\cos \theta + \sin \theta)^2(\cos^2 \theta - \cos \theta \sin \theta + \sin^2 \theta)^2$
$= (1+ \sin2\theta)(1-\sin2\theta + \sin^2 \theta \cos^2 \theta)$
$= (1+ \sin2\theta)(1-\sin2\theta) + (1 + \sin2\theta)\sin^2\theta \cos^2 \theta$
$= \cos^2 2\theta + \frac 14 (1 + \sin2\theta)\sin^2 2\theta$
$= \frac 14 (1 + \sin2\theta)\sin^2 2\theta$
Now since $\cos2\theta=0$, $\sin2\theta = \pm 1$.
Therefore the above expression can take the values 0 and $\frac12$.
My textbook gives the answer as $\frac12$. I don't see any grounds on rejecting the other answer of 0. Have I made a mistake somewhere? Or am I forgetting something?
 [/INST] You are right indeed by direct calculation we obtain
$$\cos 2\theta=0\iff \theta=\frac{\pi}4+k\frac{\pi}2$$
and since
$$\Delta^2=\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2= (-\cos^3 \theta - \sin^3 \theta)^2$$


*

*for $\theta=\frac{\pi}4 \implies \Delta^2=\left(-\frac{2\sqrt 2}{4}\right)^2=\frac12$

*for $\theta=\frac{3\pi}4\implies \Delta^2=0$

*for $\theta=\frac{5\pi}4\implies \Delta^2=\left(\frac{2\sqrt 2}{4}\right)^2=\frac12$

*for $\theta=\frac{7\pi}4\implies \Delta^2=0$

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