Trigonometric Functions – Compound Angles – JEE Main 23 Jan 2026 Shift 2

Solution:

Let the given expression be $E$. Expanding the terms:
$$ E = \sin\frac{15\theta}{2}\cos8\theta + \sin\frac{15\theta}{2}\sin8\theta + \cos\frac{15\theta}{2}\cos8\theta – \cos\frac{15\theta}{2}\sin8\theta $$

Rearranging the terms to form compound angle formulas:
$$ E = \left(\sin\frac{15\theta}{2}\cos8\theta – \cos\frac{15\theta}{2}\sin8\theta\right) + \left(\cos\frac{15\theta}{2}\cos8\theta + \sin\frac{15\theta}{2}\sin8\theta\right) $$

Using $\sin(A-B)$ and $\cos(A-B)$ formulas:
$$ E = \sin\left(\frac{15\theta}{2} – 8\theta\right) + \cos\left(\frac{15\theta}{2} – 8\theta\right) $$

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

Given $\cot\theta = -\frac{1}{2\sqrt{2}}$ and $\theta \in (\frac{\pi}{2}, \pi)$ (2nd Quadrant).


$$ \therefore \sin\theta = \frac{2\sqrt{2}}{3} $$

We need to find value of $\cos\frac{\theta}{2} – \sin\frac{\theta}{2}$. Squaring it:
$$ \left(\cos\frac{\theta}{2} – \sin\frac{\theta}{2}\right)^2 = 1 – \sin\theta = 1 – \frac{2\sqrt{2}}{3} = \frac{3-2\sqrt{2}}{3} $$

$$ = \frac{(\sqrt{2}-1)^2}{3} $$

Since $\frac{\pi}{2} < \theta < \pi \implies \frac{\pi}{4} < \frac{\theta}{2} < \frac{\pi}{2}$. In this interval, $\sin\frac{\theta}{2} > \cos\frac{\theta}{2}$, so the expression must be negative.
$$ \cos\frac{\theta}{2} – \sin\frac{\theta}{2} = -\frac{\sqrt{2}-1}{\sqrt{3}} = \frac{1-\sqrt{2}}{\sqrt{3}} $$

Ans. (1)