JEE Advanced 2024 · Paper 1, Mathematics question 3

Simplify. Regroup the expression:
$$\left(\cos\tfrac{11x}{2}\cos 6x + \sin\tfrac{11x}{2}\sin 6x\right) + \left(\sin 6x\cos\tfrac{11x}{2} - \cos 6x\sin\tfrac{11x}{2}\right) = \cos\tfrac{x}{2} + \sin\tfrac{x}{2}.$$

Values. With $\cot x = -\frac{5}{\sqrt{11}}$ and $x$ in the second quadrant, the hypotenuse is $\sqrt{25+11} = 6$, so $\cos x = -\frac56$. Since $\frac{x}{2}\in\left(\frac{\pi}{4},\frac{\pi}{2}\right)$ both half-angle ratios are positive:
$$\cos\tfrac{x}{2} = \sqrt{\tfrac{1+\cos x}{2}} = \tfrac{1}{2\sqrt3},\qquad \sin\tfrac{x}{2} = \sqrt{\tfrac{1-\cos x}{2}} = \tfrac{\sqrt{11}}{2\sqrt3}.$$

Sum $= \dfrac{\sqrt{11}+1}{2\sqrt3}$.