JEE Main 2026 Jan 23 Shift 2, Mathematics Q2: Trigonometric Ratios And Identities

Group the expression as a sum of sine and cosine addition formulas:

$\left[\sin \frac{15\theta }{2}\cos 8\theta -\cos \frac{15\theta }{2}\sin 8\theta \right]+\left[\cos \frac{15\theta }{2}\cos 8\theta +\sin \frac{15\theta }{2}\sin 8\theta \right]$

$=\sin \left(\frac{15\theta }{2}-8\theta \right)+\cos \left(\frac{15\theta }{2}-8\theta \right)$

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

Since $\frac{\pi }{2}<\theta <\pi$, $\frac{\theta }{2}\in (\frac{\pi }{4},\frac{\pi }{2})$, so $\cos \frac{\theta }{2}-\sin \frac{\theta }{2}<0$ and equals $-\sqrt{1-\sin \theta }$.

From $\cot \theta =-\frac{1}{2\sqrt{2}}$ with $\theta$ in QII, $\sin \theta =\frac{2\sqrt{2}}{3}$.

$\therefore \text{value}=-\sqrt{1-\frac{2\sqrt{2}}{3}}=-\sqrt{\frac{3-2\sqrt{2}}{3}}=-\frac{\sqrt{2}-1}{\sqrt{3}}=\frac{1-\sqrt{2}}{\sqrt{3}}$.