Trigonometry – Compound Angles – JEE Main 24 Jan 2026 Shift 1

Solution:

Let the given expression be $E$.
$$E = \sin 7x \cos \frac{13x}{2} + \sin 7x \sin \frac{13x}{2} + \cos 7x \cos \frac{13x}{2} – \cos 7x \sin \frac{13x}{2}$$
Rearranging the terms:
$$E = (\sin 7x \cos \frac{13x}{2} – \cos 7x \sin \frac{13x}{2}) + (\cos 7x \cos \frac{13x}{2} + \sin 7x \sin \frac{13x}{2})$$
Using compound angle formulas $\sin(A-B)$ and $\cos(A-B)$:
$$E = \sin(7x – \frac{13x}{2}) + \cos(7x – \frac{13x}{2})$$
$$E = \sin(\frac{x}{2}) + \cos(\frac{x}{2})$$

Given $\cot x = \frac{5}{12}$ and $x \in (\pi, \frac{3\pi}{2})$ (3rd Quadrant).
$\cos x = -\frac{5}{13}$.
We need to find half-angle values. Since $x \in (\pi, \frac{3\pi}{2})$, then $\frac{x}{2} \in (\frac{\pi}{2}, \frac{3\pi}{4})$.
In this range, $\sin \frac{x}{2} > 0$ and $\cos \frac{x}{2} < 0$.
Using $\cos x = 2\cos^2 \frac{x}{2} – 1$:
$$-\frac{5}{13} = 2\cos^2 \frac{x}{2} – 1$$
$$2\cos^2 \frac{x}{2} = 1 – \frac{5}{13} = \frac{8}{13}$$
$$\cos^2 \frac{x}{2} = \frac{4}{13} \Rightarrow \cos \frac{x}{2} = -\frac{2}{\sqrt{13}} \quad (\text{as } \frac{x}{2} \text{ is in 2nd Quad})$$

Using $\cos x = 1 – 2\sin^2 \frac{x}{2}$:
$$-\frac{5}{13} = 1 – 2\sin^2 \frac{x}{2}$$
$$2\sin^2 \frac{x}{2} = 1 + \frac{5}{13} = \frac{18}{13}$$
$$\sin^2 \frac{x}{2} = \frac{9}{13} \Rightarrow \sin \frac{x}{2} = \frac{3}{\sqrt{13}} \quad (\text{as } \frac{x}{2} \text{ is in 2nd Quad})$$

Substitute these into $E$:
$$E = \frac{3}{\sqrt{13}} + \left(-\frac{2}{\sqrt{13}}\right) = \frac{1}{\sqrt{13}}$$

Ans. (3)