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

To solve the problem, we need to simplify the given expression using trigonometric identities and the given value of \(\cot x\).

1. Given information:
\[ \frac{\pi}{2} < x < \pi, \quad \cot x = \frac{-5}{\sqrt{11}} \] Expression to simplify: \[ (\sin \frac{11x}{2})(\sin 6x - \cos 6x) + (\cos \frac{11x}{2})(\sin 6x + \cos 6x) \]

2. Group terms involving \(\sin 6x\) and \(\cos 6x\):
\[ = \sin \frac{11x}{2} \cdot \sin 6x - \sin \frac{11x}{2} \cdot \cos 6x + \cos \frac{11x}{2} \cdot \sin 6x + \cos \frac{11x}{2} \cdot \cos 6x \] Group as: \[ = (\sin \frac{11x}{2} \sin 6x + \cos \frac{11x}{2} \cos 6x) + (\cos \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x) \]

3. Use cosine and sine addition formulas:
Recall:
\[ \cos (A - B) = \cos A \cos B + \sin A \sin B \] \[ \sin (A - B) = \sin A \cos B - \cos A \sin B \] So, \[ \sin \frac{11x}{2} \sin 6x + \cos \frac{11x}{2} \cos 6x = \cos \left( \frac{11x}{2} - 6x \right) \] and \[ \cos \frac{11x}{2} \sin 6x - \sin \frac{11x}{2} \cos 6x = \sin \left( 6x - \frac{11x}{2} \right) \]

4. Simplify the angles:
\[ \frac{11x}{2} - 6x = \frac{11x}{2} - \frac{12x}{2} = -\frac{x}{2} \] \[ 6x - \frac{11x}{2} = \frac{12x}{2} - \frac{11x}{2} = \frac{x}{2} \] Therefore, the expression becomes: \[ \cos \left( -\frac{x}{2} \right) + \sin \left( \frac{x}{2} \right) \]

5. Use even-odd properties of trig functions:
\[ \cos(-\theta) = \cos \theta, \quad \sin \theta = \sin \theta \] So, \[ \cos \left( -\frac{x}{2} \right) + \sin \left( \frac{x}{2} \right) = \cos \frac{x}{2} + \sin \frac{x}{2} \]

6. Express \(\cos \frac{x}{2} + \sin \frac{x}{2}\) in a single trigonometric function:
\[ \cos \frac{x}{2} + \sin \frac{x}{2} = \sqrt{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \]

7. Find \(\sin \left( \frac{x}{2} + \frac{\pi}{4} \right)\) using the value of \(\cot x\):
Given: \[ \cot x = \frac{-5}{\sqrt{11}} = \frac{\cos x}{\sin x} \] Since \(\frac{\pi}{2} < x < \pi\), \(\sin x > 0\) and \(\cos x < 0\). Let's find \(\sin x\) and \(\cos x\):

8. Calculate \(\sin x\) and \(\cos x\):
From \(\cot x = \frac{\cos x}{\sin x} = \frac{-5}{\sqrt{11}}\), let: \[ \cos x = -5k, \quad \sin x = \sqrt{11} k, \quad k > 0 \] Using Pythagoras: \[ \sin^2 x + \cos^2 x = 1 \] \[ ( \sqrt{11} k )^2 + (-5k)^2 = 1 \] \[ 11 k^2 + 25 k^2 = 1 \implies 36 k^2 = 1 \implies k = \frac{1}{6} \] So, \[ \sin x = \frac{\sqrt{11}}{6}, \quad \cos x = -\frac{5}{6} \]

9. Use half-angle formulas:
\[ \sin \frac{x}{2} = \sqrt{ \frac{1 - \cos x}{2} } = \sqrt{ \frac{1 - \left(-\frac{5}{6}\right)}{2} } = \sqrt{ \frac{1 + \frac{5}{6}}{2} } = \sqrt{ \frac{\frac{11}{6}}{2} } = \sqrt{\frac{11}{12}} = \frac{\sqrt{11}}{2 \sqrt{3}} \] \[ \cos \frac{x}{2} = \sqrt{ \frac{1 + \cos x}{2} } = \sqrt{ \frac{1 - \frac{5}{6}}{2} } = \sqrt{ \frac{\frac{1}{6}}{2} } = \sqrt{\frac{1}{12}} = \frac{1}{2 \sqrt{3}} \]

10. Calculate \(\sin \left( \frac{x}{2} + \frac{\pi}{4} \right)\):
Using sum formula:
\[ \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) = \sin \frac{x}{2} \cos \frac{\pi}{4} + \cos \frac{x}{2} \sin \frac{\pi}{4} \] \[ = \sin \frac{x}{2} \cdot \frac{\sqrt{2}}{2} + \cos \frac{x}{2} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \left( \sin \frac{x}{2} + \cos \frac{x}{2} \right) \] \[ = \frac{\sqrt{2}}{2} \left( \frac{\sqrt{11}}{2 \sqrt{3}} + \frac{1}{2 \sqrt{3}} \right) = \frac{\sqrt{2}}{2} \times \frac{\sqrt{11} + 1}{2 \sqrt{3}} = \frac{\sqrt{2} (\sqrt{11} + 1)}{4 \sqrt{3}} \]

11. Substitute back:
\[ \cos \frac{x}{2} + \sin \frac{x}{2} = \sqrt{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) = \sqrt{2} \times \frac{\sqrt{2} (\sqrt{11} + 1)}{4 \sqrt{3}} = \frac{2 (\sqrt{11} + 1)}{4 \sqrt{3}} = \frac{\sqrt{11} + 1}{2 \sqrt{3}} \]

Final Answer:
The value of the given expression is: \[ \boxed{ \frac{\sqrt{11} + 1}{2 \sqrt{3}} } \]

Answer 2

To solve the problem, we need to evaluate the given expression based on the provided value of \(\cot x\).

1. Understanding the Given Information:
We are given that \( \cot x = -\frac{5}{\sqrt{11}} \). We need to find the value of the expression:

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

2. Using Trigonometric Identities:
Let's break the given expression into two parts. We can simplify the second part using the sum and difference trigonometric identities.

The expression becomes:

\[ \sin\left(\frac{11x}{2}\right) (\sin 6x - \cos 6x) + \cos\left(\frac{11x}{2}\right) (\sin 6x + \cos 6x) \] Using the sum formula \( \sin A \cos B + \cos A \sin B = \sin(A + B) \), we have:

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

3. Simplifying the Angle:
Now, simplify the angle of the sine term:

\[ \frac{11x}{2} + 6x = \frac{11x}{2} + \frac{12x}{2} = \frac{23x}{2} \] So, the expression simplifies to:

\[ \sin\left(\frac{23x}{2}\right) \]

4. Relating \( \cot x \) to \( \sin x \) and \( \cos x \):
We are given that \( \cot x = -\frac{5}{\sqrt{11}} \), which means:

\[ \cot x = \frac{\cos x}{\sin x} \] Thus, \( \cos x = -\frac{5}{\sqrt{11}} \sin x \). Now, we can use the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) to find \( \sin x \) and \( \cos x \):

\[ \cos^2 x = \left( -\frac{5}{\sqrt{11}} \sin x \right)^2 = \frac{25}{11} \sin^2 x \] Substituting into the identity:

\[ \sin^2 x + \frac{25}{11} \sin^2 x = 1 \] Factor out \( \sin^2 x \):

\[ \sin^2 x \left( 1 + \frac{25}{11} \right) = 1 \] Simplify:

\[ \sin^2 x \times \frac{36}{11} = 1 \] Solving for \( \sin^2 x \):

\[ \sin^2 x = \frac{11}{36} \] Thus:

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

5. Substituting Values into the Expression:
Now that we have the values of \( \sin x \) and \( \cos x \), we can substitute these into the original expression \( \sin\left(\frac{23x}{2}\right) \). Using trigonometric identities, the expression evaluates to:

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

Final Answer:
The correct answer is \( \boxed{(B)} \).