\(sin[2sin^{-1}\frac{4}{5}]\) = ?A. \(\frac{12}{25}\)B. \(\frac{16}

1.	\(sin[2sin^{-1}\frac{4}{5}]\) = ?A. \(\frac{12}{25}\)B. \(\frac{16}{25}\)C. \(\frac{24}{25}\)D. none of these
Answer» Correct Answer is (C) \(\frac{24}{25}\) Let, x = \(sin^{-1}\frac{4}{5}\) ⇒ sin x = \(\frac{4}{5}\) We know that ,cos x =\(\sqrt{1-sin^2x}\) \(=\sqrt{1-(\frac{4}{5})^2}\) = \(\frac{3}{5}\) Now since, x = \(sin^{-1}\frac{4}{5}\), hence \(sin(2sin^{-1}\frac{4}{5})\) become sin (2x) Here, sin (2x) = 2 sin x cos x = 2 x \(\frac{4}{5}\) x \(\frac{3}{5}\) = \(\frac{24}{25}\)

\(sin[2sin^{-1}\frac{4}{5}]\) = ?A. \(\frac{12}{25}\)B. \(\frac{16}{25}\)C. \(\frac{24}{25}\)D. none of these

Answer»

Correct Answer is (C) \(\frac{24}{25}\)

Let, x = \(sin^{-1}\frac{4}{5}\)

⇒ sin x = \(\frac{4}{5}\)

We know that ,cos x =\(\sqrt{1-sin^2x}\)

\(=\sqrt{1-(\frac{4}{5})^2}\)

= \(\frac{3}{5}\)

Now since, x = \(sin^{-1}\frac{4}{5}\), hence \(sin(2sin^{-1}\frac{4}{5})\) become sin (2x)

Here, sin (2x) = 2 sin x cos x

= 2 x \(\frac{4}{5}\) x \(\frac{3}{5}\)

= \(\frac{24}{25}\)