Let (π)/(2) ((15 θ)/(2))( 8 θ+ 8 θ)+ ((15 θ)/(2))( 8 θ- 8 θ) is...

Expanding and regrouping:
$\sin\frac{15\theta}{2}(\cos 8\theta + \sin 8\theta) + \cos\frac{15\theta}{2}(\cos 8\theta - \sin 8\theta)$
$= \left(\cos\frac{15\theta}{2}\cos 8\theta + \sin\frac{15\theta}{2}\sin 8\theta\right) + \left(\sin\frac{15\theta}{2}\cos 8\theta - \cos\frac{15\theta}{2}\sin 8\theta\right)$
$= \cos\left(8\theta - \frac{15\theta}{2}\right) + \sin\left(\frac{15\theta}{2} - 8\theta\right) = \cos\frac{\theta}{2} - \sin\frac{\theta}{2}$.
From $\cot\theta = -\frac{1}{2\sqrt{2}}$: $\sin\theta = \frac{2\sqrt{2}}{3}$, $\cos\theta = -\frac{1}{3}$.
\left(\cos\frac{\theta}{2} - \sin\frac{\theta}{2}\right)^2 = 1 - \sin\theta = 1 - \frac{2\sqrt{2}}{3} = \frac{(\sqrt{2}-1)^2}{3}$.<br>Since$\frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{2}$,$\sin\frac{\theta}{2} > \cos\frac{\theta}{2}$, so the value is$\frac{1-\sqrt{2}}{\sqrt{3}}.