Let $\frac{\pi}{2} < x < \pi$ be such that $\cot x = \frac{-5}{\sqrt{11}}$. Then $\left(\sin\frac{11x}{2}\right)(\sin 6x - \cos 6x) + \left(\cos\frac{11x}{2}\right)(\sin 6x + \cos 6x)$ is equal to

Let π/2 < x < π be such that cot x = (-5)/√11. Then (sin11x/2)(sin 6x - cos 6x) + (cos11x/2)(sin 6x + cos 6x) is equal to

$\frac{\pi}{2} < x < \pi \quad (\text{second quadrant})$

$\cot x = \frac{-5}{\sqrt{11}}$

$E = \sin\frac{11x}{2}(\sin 6x - \cos 6x) + \cos\frac{11x}{2}(\sin 6x + \cos 6x)$

$\text{Plan: simplify } E \text{ with angle-sum formulas, then use half-angles}$

$\cot x = -\frac{5}{\sqrt{11}} \ \Rightarrow\ \tan x = -\frac{\sqrt{11}}{5}$

$\text{hyp} = \sqrt{5^{2} + (\sqrt{11})^{2}} = \sqrt{36} = 6$

$\sin x = \frac{\sqrt{11}}{6}, \quad \cos x = -\frac{5}{6}$

$E = \sin\frac{11x}{2}\sin 6x - \sin\frac{11x}{2}\cos 6x + \cos\frac{11x}{2}\sin 6x + \cos\frac{11x}{2}\cos 6x$

$E = \left(\sin\frac{11x}{2}\sin 6x + \cos\frac{11x}{2}\cos 6x\right) + \left(\cos\frac{11x}{2}\sin 6x - \sin\frac{11x}{2}\cos 6x\right)$

$= \cos\!\left(\frac{11x}{2} - 6x\right) + \sin\!\left(6x - \frac{11x}{2}\right)$

$= \cos\frac{x}{2} + \sin\frac{x}{2}$

$E = \sin\frac{x}{2} + \cos\frac{x}{2}$

$\sin^2\frac{x}{2} = \frac{1-\cos x}{2}, \quad \cos^2\frac{x}{2} = \frac{1+\cos x}{2}$

$\sin^2\frac{x}{2} = \frac{1+\frac{5}{6}}{2} = \frac{11}{12}$

$\cos^2\frac{x}{2} = \frac{1-\frac{5}{6}}{2} = \frac{1}{12}$

$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \ \Rightarrow\ \sin\frac{x}{2}, \cos\frac{x}{2} > 0$

$\sin\frac{x}{2} = \sqrt{\frac{11}{12}} = \frac{\sqrt{11}}{2\sqrt{3}}, \quad \cos\frac{x}{2} = \sqrt{\frac{1}{12}} = \frac{1}{2\sqrt{3}}$

$E = \frac{1}{2\sqrt{3}} + \frac{\sqrt{11}}{2\sqrt{3}} = \frac{\sqrt{11}+1}{2\sqrt{3}}$

$\boxed{E = \frac{\sqrt{11}+1}{2\sqrt{3}}} \Rightarrow \text{option (2)}$