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?

When in doubt, use the relations in the original problem. Let $\theta=\frac{3\pi}4$. Then $\cos2\theta=0$, while $\sin\theta=\sqrt2/2=a$ and $\cos\theta=-\sqrt2/2=-a$.

The expression in the question is now $$\begin{vmatrix}0&-a&a\\-a&a&0\\a&0&-a\end{vmatrix}^2$$ Clearly, adding up all three rows of the matrix produces the zero vector, so the whole expression evaluates to zero. Your working is entirely correct: the book is wrong to omit 0 as an answer.

The result of $\frac12$ is obtained with the other principal solution to $\cos2\theta=0$, $\theta=\frac\pi4$.

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

If we compute the square of the matrix, we get $$\begin{bmatrix} 1 & \sin\theta\cos\theta & \sin\theta\cos\theta \\ \sin\theta\cos\theta & 1 & \sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin\theta\cos\theta & 1 \end{bmatrix}= \frac{1}{2} \begin{bmatrix} 2 & \sin2\theta & \sin2\theta \\ \sin2\theta & 2 & \sin2\theta \\ \sin2\theta & \sin2\theta & 2 \end{bmatrix}$$ whose determinant is $$\frac{1}{8}(8+2\sin^32\theta-6\sin^22\theta)$$ From $\cos2\theta=0$, we have $\sin^22\theta=1$, so finally we get $$\Delta^2=\frac{1}{4}(1+\sin2\theta)= \begin{cases} 1/2 & \text{if \sin2\theta=1}\\[4px] 0 & \text{if \sin2\theta=-1} \end{cases}$$ both possibilities being allowed by the hypothesis that $\cos2\theta=0$.

• Yeah, I tried this too but still got 2 answers. So I thought the book might be wrong. – SamInuyasha ANMF Mar 2 '18 at 7:09
• @SamInuyashaANMF Indeed it is wrong, unless the matrix is not assumed to have further properties, such as being invertible. – egreg Mar 2 '18 at 7:13