JEE Advanced 2024 Paper 1 Q03 Mathematics Trigonometry Trigonometric Identities Medium

JEE Advanced 2024 Paper 1 · Q03 · Trigonometric Identities

Let $\dfrac{\pi}{2} < x < \pi$ be such that $\cot x = \dfrac{-5}{\sqrt{11}}$. Then $$\left(\sin \dfrac{11x}{2}\right)(\sin 6x - \cos 6x) + \left(\cos \dfrac{11x}{2}\right)(\sin 6x + \cos 6x)$$ is equal to

  1. A. $\dfrac{\sqrt{11} - 1}{2\sqrt{3}}$
  2. B. $\dfrac{\sqrt{11} + 1}{2\sqrt{3}}$
  3. C. $\dfrac{\sqrt{11} + 1}{3\sqrt{2}}$
  4. D. $\dfrac{\sqrt{11} - 1}{3\sqrt{2}}$
Reveal answer + step-by-step solution

Correct answer:B

Solution

Group the expression as $\sin\frac{11x}{2}\sin 6x + \cos\frac{11x}{2}\cos 6x + \cos\frac{11x}{2}\sin 6x - \sin\frac{11x}{2}\cos 6x = \cos\!\left(6x-\frac{11x}{2}\right) + \sin\!\left(6x-\frac{11x}{2}\right) = \cos\frac{x}{2}+\sin\frac{x}{2}$. Since $\cot x=\dfrac{-5}{\sqrt{11}}$ with $x$ in QII, $\cos x=-\dfrac{5}{6}$ (using $\csc^2 x=1+\cot^2 x=\frac{36}{11}$, $\sin x=\frac{\sqrt{11}}{6}$). Then $\cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}=\sqrt{\frac{1/6}{2}}=\frac{1}{2\sqrt{3}}$ and $\sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}=\sqrt{\frac{11/6}{2}}=\frac{\sqrt{11}}{2\sqrt{3}}$. Sum $=\dfrac{\sqrt{11}+1}{2\sqrt{3}}$, option (B).

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