Trigonometry & Inverse Trigonometry
Trigonometry
star_batch_jee_advanced_2025
Grade 11
Question:
If $A, B, C, D$ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $k$, then the value of $4\sin\frac{A}{2}+3\sin\frac{B}{2}+2\sin\frac{C}{2}+\sin\frac{D}{2}$ is equal to:
$2\sqrt{1-k}$
$\sqrt{1+k}$
$2\sqrt{k}$
None of these
Step-by-Step Solution
Key Concept: Strategic choice of angles using periodicity and symmetry of sine function reduces the expression to a form involving $\sqrt{1 + k}$.
Given conditions $A < B < C < D$ and $\sin A = \sin B = \sin C = \sin D = k$. Choosing $B = \pi - A$, $C = 2\pi + A$, $D = 3\pi - A$ satisfies the conditions. Computing $4\sin\frac{A}{2} + 3\sin\frac{\pi-A}{2} + 2\sin\frac{2\pi+A}{2} + \sin\frac{3\pi-A}{2}$ simplifies to $2\left(\sin^2\frac{A}{2} + \cos^2\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}\right) = 2\sqrt{1 + \sin A} = 2\sqrt{1 + k}$.
Correct Answer: 4