Let \(\frac{\pi }{2}

Explanation

1. Simplify the trigonometric expression

Let the given expression be \(E\).

\(E=(\sin ⁡\frac{11x}{2})(\sin ⁡6x−\cos ⁡6x)+(\cos ⁡\frac{11x}{2})(\sin ⁡6x+\cos ⁡6x)\)

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:

\(E=(\cos ⁡\frac{11x}{2}\cos ⁡6x+\sin ⁡\frac{11x}{2}\sin ⁡6x)+(\cos ⁡\frac{11x}{2}\sin ⁡6x−\sin ⁡\frac{11x}{2}\cos ⁡6x)\)

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 \(\cos ⁡(\frac{11x}{2}−6x)\).

The second part is \(\sin ⁡(6x−\frac{11x}{2})\).

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:

\(E=\cos ⁡(−\frac{x}{2})+\sin ⁡(\frac{x}{2})\)

Since cosine is an even function

, \(\cos ⁡(−\theta )=\cos ⁡(\theta )\):

\(E=\cos ⁡(\frac{x}{2})+\sin ⁡(\frac{x}{2})\)

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}

The condition \(\frac{\pi }{2}

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 \(\sin ⁡(\frac{x}{2})\) and \(\cos ⁡(\frac{x}{2})\)

Given \(\frac{\pi }{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 \(\sin ⁡(\frac{x}{2})\) and

\(\cos ⁡(\frac{x}{2})\) are positive.

We use the half-angle formulas:

\({\sin ⁡}^{2}(\frac{x}{2})=\frac{1−\cos ⁡x}{2}\)

\({\cos ⁡}^{2}(\frac{x}{2})=\frac{1+\cos ⁡x}{2}\)

Substitute the value of \(\cos ⁡x=−\frac{5}{6}\):

\({\sin ⁡}^{2}(\frac{x}{2})=\frac{1−(−\frac{5}{6})}{2}=\frac{1+\frac{5}{6}}{2}=\frac{\frac{6+5}{6}}{2}=\frac{\frac{11}{6}}{2}=\frac{11}{12}\)

Since \(\sin ⁡(\frac{x}{2})>0\):

\(\sin ⁡(\frac{x}{2})=\sqrt{\frac{11}{12}}=\frac{\sqrt{11}}{\sqrt{4⋅3}}=\frac{\sqrt{11}}{2\sqrt{3}}\)

And for \(\cos ⁡(\frac{x}{2})\):

\({\cos ⁡}^{2}(\frac{x}{2})=\frac{1+(−\frac{5}{6})}{2}=\frac{1−\frac{5}{6}}{2}=\frac{\frac{6−5}{6}}{2}=\frac{\frac{1}{6}}{2}=\frac{1}{12}\)

Since \(\cos ⁡(\frac{x}{2})>0\):

\(\cos ⁡(\frac{x}{2})=\sqrt{\frac{1}{12}}=\frac{1}{\sqrt{4⋅3}}=\frac{1}{2\sqrt{3}}\)

1. Calculate the final expression

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

into the simplified expression \(E\):

\(E=\cos ⁡(\frac{x}{2})+\sin ⁡(\frac{x}{2})\)

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

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

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 \(\text{B }\frac{\sqrt{11}+1}{2\sqrt{3}}\).