Q5

If one computes the above 3x3 determinant, the result is $D(\theta) = 2 + \sin2\theta + \cos2\theta$ . The range of $D$ can be determined via its first & second derivatives:

$D'(\theta) = 0 \Rightarrow 2\cos2\theta - 2\sin2\theta = 0 \Rightarrow \tan2\theta = 1 \Rightarrow 2\theta = \frac{\pi}{4}, \frac{3\pi}{4} \Rightarrow \theta = \frac{\pi}{8}, \frac{3\pi}{8}$ (i).

$D''(\frac{\pi}{8}) = -4\sin2(\frac{\pi}{8}) - 4\cos2(\frac{\pi}{8}) = -4(sin(\frac{\pi}{4}) + cos(\frac{\pi}{4})) = -4\sqrt{2} < 0$ (MAXIMUM) (ii)

$D''(\frac{3\pi}{8}) = -4\sin2(\frac{3\pi}{8}) - 4\cos2(\frac{3\pi}{8}) = -4(sin(\frac{3\pi}{4}) + cos(\frac{3\pi}{4})) = 4\sqrt{2} > 0$ (MINIMUM) (ii)

Finally, we obtain $D(\frac{\pi}{8}) = \boxed{2 + \sqrt{2}}$ and $D(\frac{3\pi}{8}) = \boxed{2-\sqrt{2}}.$