Question

Find the value of sin (2 sin⁻¹ $\frac{4}{5}$).

Answer 1

hope it help you dear......

Answer 2

Answer:

$sin(2\:sin^{-1}\frac{4}{5})=\frac{24}{25}$

Step-by-step explanation:

To find the value of

$sin(2\:sin^{-1}\frac{4}{5})$

As

$2\:sin^{-1}x=sin^{-1}[2x\sqrt{1-x^{2}} ]\\\\\\2\:sin^{-1}\frac{4}{5} =sin^{-1}[\frac{2\times4}{5}\sqrt{1-(\frac{4}{5})^{2}} ]\\\\\\=sin^{-1}[\frac{8}{5}\sqrt{1-(\frac{16}{25})} ]\\\\\\=sin^{-1}[\frac{8}{5}\sqrt{(\frac{25-16}{25})} ]\\\\\\=sin^{-1}[\frac{8}{5}\sqrt{(\frac{9}{25})} ]\\\\\\=sin^{-1}[\frac{8}{5}(\frac{3}{5}) ]$

$=sin^{-1}[\frac{24}{25}]$

so

$sin(2\:sin^{-1}\frac{4}{5})=sin(sin^{-1}\frac{24}{25})\\\\\\=\frac{24}{25}$