Question:
The value of $\sin\left(2\sin^{-1}\frac{3}{5}\right)$ is equal to
The value of $\sin\left(2\sin^{-1}\frac{3}{5}\right)$ is equal to
Show Hint
Convert inverse trig into triangle form to easily find other ratios.
Updated On: Apr 24, 2026
- $\frac{23}{25}$
- $\frac{21}{25}$
- $\frac{22}{25}$
- $\frac{24}{25}$
- $\frac{18}{25}$
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The Correct Option is D
Solution and Explanation
Concept:
• $\sin(2\theta) = 2\sin\theta \cos\theta$
Step 1: Let $\theta = \sin^{-1}\frac{3}{5}$
\[ \sin\theta = \frac{3}{5} \]
Step 2: Find $\cos\theta$
\[ \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} \]
Step 3: Apply identity
\[ \sin(2\theta) = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25} \] Final Conclusion:
\[ = \frac{24}{25} \]
• $\sin(2\theta) = 2\sin\theta \cos\theta$
Step 1: Let $\theta = \sin^{-1}\frac{3}{5}$
\[ \sin\theta = \frac{3}{5} \]
Step 2: Find $\cos\theta$
\[ \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5} \]
Step 3: Apply identity
\[ \sin(2\theta) = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25} \] Final Conclusion:
\[ = \frac{24}{25} \]
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