Question

Let \(\frac{\pi}{2}<x<\pi\) be such that \(\cot x=\frac{-5}{\sqrt{11}}\). Then
\((\sin\frac{11x}{2})(\sin 6x-\cos6x)+(\cos\frac{11x}{2})(\sin 6x+\cos 6x)\)
is equal to

• A. \(\frac{\sqrt{11}-1}{2\sqrt3}\)
• B. \(\frac{\sqrt{11}+1}{2\sqrt3}\)
• C. \(\frac{\sqrt{11}+1}{3\sqrt2}\)
• D. \(\frac{\sqrt{11}-1}{3\sqrt2}\)

Answer 1

The Correct Option is B

Let 

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

Simplify using trigonometric identities:

\[ E = \sin \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x + \cos \frac{11x}{2} \sin 6x + \cos \frac{11x}{2} \cos 6x. \]

Grouping terms:

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

Simplify further:

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

Squaring both sides:

\[ E^2 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2 \cos \frac{x}{2} \sin \frac{x}{2}. \] \[ E^2 = 1 + \sin x. \]

Since \( \frac{\pi}{2} < x < \pi \) and \( \sin x = \frac{\sqrt{11}}{6} \), we get:

\[ E^2 = 1 + \frac{\sqrt{11}}{6}. \] \[ E = \sqrt{\frac{11 + 1}{2 \sqrt{3}}}. \]

Final Answer:

\[ \frac{\sqrt{11} + 1}{2 \sqrt{3}} \]