JEE Advanced 2024 Paper 1, Mathematics Section 1 Q3: Trigonometric Ratios And Identities

Group the expression as a sum of two sine/cosine combinations:

$$E=\sin 6x\cos \frac{11x}{2}-\cos 6x\sin \frac{11x}{2}+\cos 6x\cos \frac{11x}{2}+\sin 6x\sin \frac{11x}{2}$$

The first pair gives $\sin \left(6x-\frac{11x}{2}\right)=\sin \frac{x}{2}$, and the second pair gives $\cos \left(6x-\frac{11x}{2}\right)=\cos \frac{x}{2}$.

$$\therefore E=\sin \frac{x}{2}+\cos \frac{x}{2}⟹E^2=1+\sin x$$

Since $\cot x=\frac{-5}{\sqrt{11}}$ and $x\in \left(\frac{\pi }{2},\pi \right)$, we have $\sin x>0$. From the right triangle with legs $\sqrt{11}$ and $5$, hypotenuse $\sqrt{36}=6$, so $\sin x=\frac{\sqrt{11}}{6}$.

$$E^2=1+\frac{\sqrt{11}}{6}=\frac{6+\sqrt{11}}{6}=\frac{12+2\sqrt{11}}{12}=\frac{(\sqrt{11}+1{)}^2}{12}$$

Also, on this interval $\frac{x}{2}\in \left(\frac{\pi }{4},\frac{\pi }{2}\right)$, so both $\sin \frac{x}{2}$ and $\cos \frac{x}{2}$ are positive, giving $E>0$.

$$\therefore E=\frac{\sqrt{11}+1}{2\sqrt{3}}$$