Question

Show that $2{\sin }^{-1}\left(\frac{4}{5}\right)={\tan }^{-1}\left(\frac{24}{7}\right)$

Answer 1

$\theta =2{\sin }^{-1}\left(\frac{4}{5}\right)$

$\frac{\theta }{2}={\sin }^{-1}\left(\frac{4}{5}\right)$

$\sin \frac{\theta }{2}=\frac{4}{5}$

(Image 1)

$\cos \frac{\theta }{2}=\frac{3}{5}$

$\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$

$=2\times \frac{4}{5}\times \frac{3}{5}=\frac{24}{25}$

(Image 2)

$\sin \theta =\frac{24}{25}$

${AB}^2+{BC}^2={AC}^2$

$BC=\sqrt{{AC}^2-{AB}^2}$

$BC=7$

$\tan \theta =\frac{24}{7}$

$\theta ={\tan }^{-1}\left(\frac{24}{7}\right)$

$\therefore \theta =2{\sin }^{-1}\left(\frac{4}{5}\right)={\tan }^{-1}\left(\frac{24}{7}\right)$