Question

If \( \cot\theta = -\dfrac{1}{2\sqrt{2}} \), where \( \theta \in \left( \dfrac{3\pi}{2}, 2\pi \right) \), then the value of \[ \sin\left(\dfrac{15\theta}{2}\right)(\sin 8\theta + \cos 8\theta) + \cos\left(\dfrac{15\theta}{2}\right)(\cos 8\theta - \sin 8\theta) \] is:

• A. \( \dfrac{\sqrt{2}+1}{\sqrt{3}} \)
• B. \( -\dfrac{\sqrt{2}+1}{\sqrt{3}} \)
• C. \( \dfrac{\sqrt{2}-1}{\sqrt{3}} \)
• D. \( -\dfrac{\sqrt{2}-1}{\sqrt{3}} \)

Hint:

Expressions involving mixed sine and cosine terms often reduce using sum-to-product or cosine difference identities.

Answer 1

The Correct Option is B

Step 1: Use trigonometric identity.
The given expression is of the form \[ \sin A(\sin B + \cos B) + \cos A(\cos B - \sin B) \] which simplifies to \[ \cos(B - A) \]
Step 2: Apply the identity.
Here, \[ A = \dfrac{15\theta}{2}, \quad B = 8\theta \] So the expression becomes \[ \cos\left(8\theta - \dfrac{15\theta}{2}\right) = \cos\left(\dfrac{\theta}{2}\right) \]
Step 3: Find \( \cos\dfrac{\theta}{2} \).
Given \( \cot\theta = -\dfrac{1}{2\sqrt{2}} \) and \( \theta \in \left(\dfrac{3\pi}{2}, 2\pi\right) \), we get \[ \sin\theta<0,\quad \cos\theta>0 \] Using right triangle ratios, \[ \cos\dfrac{\theta}{2} = -\dfrac{\sqrt{2}+1}{\sqrt{3}} \]
Step 4: Final Answer.
\[ \boxed{-\dfrac{\sqrt{2}+1}{\sqrt{3}}} \]