Evaluate $displaystylesin\left(2si{n}^{-1}\frac{3}{5}\right)$.

Step-by-step solution

1. Let $\theta =si{n}^{-1}\frac{3}{5}$ so $sin\theta =\frac{3}{5}$ and $\theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$.
2. Use $sin(2\theta )=2sin\theta \cos \theta$. We need $cos\theta =\sqrt{1-si{n}^2\theta }=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$. Thus $sin(2\theta )=2\times \frac{3}{5}\times \frac{4}{5}=\frac{24}{25}$.
3. Hence the correct option is (b).

Concept tested: Double–angle for sine of arcsin

Published 18 September 2026 · How this solution was prepared and checked