Trigonometric Functions

Explanation

1. Simplify the trigonometric expression

Let the given expression be $E$.

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

Expand the terms:

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

Rearrange the terms to group familiar identities:

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

We recognize the cosine difference formula

$\cos(A-B) = \cos A \cos B + \sin A \sin B$ and

the sine difference formula

$\sin(B-A) = \sin B \cos A - \cos B \sin A$.

Let $A = \frac{11x}{2}$ and $B = 6x$.

The first part is $\cos\left(\frac{11x}{2} - 6x\right)$.

The second part is $\sin\left(6x - \frac{11x}{2}\right)$.

Calculate the arguments:

$A - B = \frac{11x}{2} - \frac{12x}{2} = -\frac{x}{2}$

$B - A = \frac{12x}{2} - \frac{11x}{2} = \frac{x}{2}$

Substitute these back into the expression:

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

Since cosine is an even function

, $\cos(-\theta) = \cos(\theta)$:

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

1. Determine $\sin x$ and $\cos x$ from the given $\cot x$

We are given $\cot x = -\frac{5}{\sqrt{11}}$ and $\frac{\pi}{2} < x < \pi$.

The condition $\frac{\pi}{2} < x < \pi$ means $x$ is

in the second quadrant. In the second quadrant,

$\sin x > 0$ and $\cos x < 0$.

We can construct a right triangle where

the adjacent side is 5 and the opposite

side is $\sqrt{11}$ (ignoring the sign for now).

The hypotenuse $h$ is calculated

using the Pythagorean theorem:

$h = \sqrt{5^2 + (\sqrt{11})^2} = \sqrt{25 + 11} = \sqrt{36} = 6$

Now, we can find $\sin x$ and $\cos x$:

$\sin x = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{11}}{6}$

$\cos x = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{6}$

Applying the signs for the second quadrant:

$\sin x = \frac{\sqrt{11}}{6}$

$\cos x = -\frac{5}{6}$

1. Determine $\sin\left(\frac{x}{2}\right)$ and $\cos\left(\frac{x}{2}\right)$

Given $\frac{\pi}{2} < x < \pi$, we can find the range for $\frac{x}{2}$:

$\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$

This means $\frac{x}{2}$ is in the first quadrant.

In the first quadrant, both $\sin\left(\frac{x}{2}\right)$ and

$\cos\left(\frac{x}{2}\right)$ are positive.

We use the half-angle formulas:

$\sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos x}{2}$

$\cos^2\left(\frac{x}{2}\right) = \frac{1 + \cos x}{2}$

Substitute the value of $\cos x = -\frac{5}{6}$:

$\sin^2\left(\frac{x}{2}\right) = \frac{1 - \left(-\frac{5}{6}\right)}{2} = \frac{1 + \frac{5}{6}}{2} = \frac{\frac{6+5}{6}}{2} = \frac{\frac{11}{6}}{2} = \frac{11}{12}$

Since $\sin\left(\frac{x}{2}\right) > 0$:

$\sin\left(\frac{x}{2}\right) = \sqrt{\frac{11}{12}} = \frac{\sqrt{11}}{\sqrt{4 \cdot 3}} = \frac{\sqrt{11}}{2\sqrt{3}}$

And for $\cos\left(\frac{x}{2}\right)$:

$\cos^2\left(\frac{x}{2}\right) = \frac{1 + \left(-\frac{5}{6}\right)}{2} = \frac{1 - \frac{5}{6}}{2} = \frac{\frac{6-5}{6}}{2} = \frac{\frac{1}{6}}{2} = \frac{1}{12}$

Since $\cos\left(\frac{x}{2}\right) > 0$:

$\cos\left(\frac{x}{2}\right) = \sqrt{\frac{1}{12}} = \frac{1}{\sqrt{4 \cdot 3}} = \frac{1}{2\sqrt{3}}$

1. Calculate the final expression

Now substitute the values of $\sin\left(\frac{x}{2}\right)$ and $\cos\left(\frac{x}{2}\right)$

into the simplified expression $E$:

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

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

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

Comparing this result with the given options:

A: $\frac{\sqrt{11}-1}{2\sqrt{3}}$

B: $\frac{\sqrt{11}+1}{2\sqrt{3}}$

C: $\frac{\sqrt{11}+1}{3\sqrt{2}}$

D: $\frac{\sqrt{11}-1}{3\sqrt{2}}$

The calculated value matches option B.

The final answer is $\boxed{\text{B } \frac{\sqrt{11}+1}{2\sqrt{3}}}$.