Question 3

Let the expression be $E$. $E = \sin \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x + \cos \frac{11x}{2} \sin 6x + \cos \frac{11x}{2} \cos 6x$ Using sum and difference formulas: $E = \left( \cos \frac{11x}{2} \cos 6x + \sin \frac{11x}{2} \sin 6x \right) + \left( \sin 6x \cos \frac{11x}{2} - \cos 6x \sin \frac{11x}{2} \right)$ $E = \cos\left(6x - \frac{11x}{2}\right) + \sin\left(6x - \frac{11x}{2}\right) = \cos\frac{x}{2} + \sin\frac{x}{2}$ Given $\cot x = -\frac{5}{\sqrt{11}}$ and $\frac{\pi}{2} < x < \pi$. $\operatorname{cosec}^2 x = 1 + \cot^2 x = 1 + \frac{25}{11} = \frac{36}{11} \implies \sin x = \frac{\sqrt{11}}{6}$ (since $x$ is in 2nd quadrant). We know $\left( \sin \frac{x}{2} + \cos \frac{x}{2} \right)^2 = 1 + \sin x = 1 + \frac{\sqrt{11}}{6} = \frac{6 + \sqrt{11}}{6}$. Since $\frac{\pi}{2} < x < \pi \implies \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$, both $\sin \frac{x}{2}$ and $\cos \frac{x}{2}$ are positive. Thus, $E = \sqrt{\frac{6 + \sqrt{11}}{6}} = \sqrt{\frac{12 + 2\sqrt{11}}{12}} = \sqrt{\frac{(\sqrt{11}+1)^2}{12}} = \frac{\sqrt{11}+1}{\sqrt{12}} = \frac{\sqrt{11}+1}{2\sqrt{3}}$.