Compute sin(2sin^-13/5).

Compute $sin (2 sin^{- 1} \frac{3}{5})$ .

hard 4 marks Class 12 · Mathematics · Inverse Trigonometric Functions

Answer

\frac{24}{25}

The correct answer is $\frac{24}{25}$ — Substitute the values: $sin (2 θ) = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$ .

Let $θ = sin^{- 1} \frac{3}{5}$ , so $sin θ = \frac{3}{5}$ .
Find $cos θ$ using $cos θ = 1 - sin^{2} θ = 1 - (\frac{3}{5})^{2} = \frac{4}{5}$ (positive because $θ \in [- \frac{π}{2}, \frac{π}{2}]$ ).
Use the double‑angle identity $sin (2 θ) = 2 sin θ cos θ$ .
Substitute the values: $sin (2 θ) = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25}$ .

Common mistake: Taking

cos θ

- \frac{4}{5}

instead of

\frac{4}{5}

Concept tested: Double‑angle formula with inverse sine