Question

Let \( \dfrac{\pi}{2} < \theta < \pi \) and \( \cot \theta = -\dfrac{1}{2\sqrt{2}} \). Then the value of \[ \sin\!\left(\frac{15\theta}{2}\right)(\cos 8\theta + \sin 8\theta) + \cos\!\left(\frac{15\theta}{2}\right)(\cos 8\theta - \sin 8\theta) \] is equal to

• A. $\frac{\sqrt{2}-1}{\sqrt{3}}$
• B. $\frac{\sqrt{2}}{\sqrt{3}}$
• C. $\frac{1-\sqrt{2}}{\sqrt{3}}$
• D. $\frac{\sqrt{2}}{\sqrt{3}}$

Hint:

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

Answer 1

The Correct Option is C

Expression: $\sin(A)\cos(B) + \sin(A)\sin(B) + \cos(A)\cos(B) - \cos(A)\sin(B)$ with $A=15\theta/2, B=8\theta$.
Group terms: $(\sin A \cos B - \cos A \sin B) + (\cos A \cos B + \sin A \sin B)$.
$= \sin(A-B) + \cos(A-B)$.
$A-B = \frac{15\theta}{2} - \frac{16\theta}{2} = -\frac{\theta}{2}$.
Exp $= \sin(-\theta/2) + \cos(-\theta/2) = \cos(\theta/2) - \sin(\theta/2)$.
Given $\cot \theta = -1/(2\sqrt{2})$, $\cos\theta = -1/3$.
$\theta$ in Q2 $\implies \theta/2$ in Q1.
$\sin(\theta/2) = \sqrt{\frac{1 - (-1/3)}{2}} = \sqrt{2/3}$.
$\cos(\theta/2) = \sqrt{\frac{1 + (-1/3)}{2}} = \sqrt{1/3}$.
Result $= \frac{1}{\sqrt{3}} - \frac{\sqrt{2}}{\sqrt{3}}$.