Question

$\frac{\pi}{2} < \theta < \pi$ and $cot\theta = - \frac{1}{2\sqrt{2}}$

$${\Rightarrow sin\left( \frac{15\theta}{2} \right)(cos8\theta + sin8\theta) }{+ cos\left( \frac{15\theta}{2} \right)(cos8\theta - sin8\theta) }{\Rightarrow sin\left( \frac{15\theta}{2} \right)cos8\theta - cos\left( \frac{15\theta}{2} \right)sin8\theta }{+ sin\frac{15\theta}{2}sin8\theta + cos\frac{15\theta}{2}cos8\theta }{\Rightarrow sin\left( \frac{15\theta}{2} - 8\theta \right) + cos\left( \frac{15\theta}{2} - 8\theta \right) }{\Rightarrow cos\frac{\theta}{2} - sin\frac{\theta}{2} = - \sqrt{1 - sin\theta}\ \left( \frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{2} \right) }$$⇒ given $cot\theta = - \frac{1}{2\sqrt{2}},sin\theta = \frac{2\sqrt{2}}{3}$

$${\Rightarrow - \sqrt{1 - sin\theta} = \sqrt{\frac{3 - 2\sqrt{2}}{3}} = - \frac{(\sqrt{2} - 1)}{3} }{= \frac{1 - \sqrt{2}}{\sqrt{3}}}$$