Question

Prove $2{\sin }^{-1}\left(\frac{5}{13}\right)={\cos }^{-1}\left(\frac{119}{169}\right)$

Answer 1

Applying the trigonometric identities in the LHS of $2{\sin }^{-1}\left(\frac{5}{13}\right)={\cos }^{-1}\left(\frac{119}{169}\right)$.

$2{\sin }^{-1}\left(\frac{5}{13}\right)={\sin }^{-1}\left(2\left(\frac{5}{13}\right)\sqrt{1-{\left(\frac{5}{13}\right)}^2}\right)$

$={\sin }^{-1}\left(\frac{10}{13}\sqrt{1-\frac{25}{169}}\right)$

$={\sin }^{-1}\left(\frac{10}{13}\sqrt{\frac{144}{169}}\right)$

$={\sin }^{-1}\left(\left(\frac{10}{13}\right)\left(\frac{12}{13}\right)\right)$

$={\sin }^{-1}\left(\frac{120}{169}\right)$

Let ${\sin }^{-1}\left(\frac{120}{169}\right)=\theta$,

$\frac{120}{169}=\sin \theta$

$\cos \theta =\frac{119}{169}$

$\theta ={\cos }^{-1}\left(\frac{119}{169}\right)$

${\sin }^{-1}\left(\frac{120}{169}\right)={\cos }^{-1}\left(\frac{119}{169}\right)$

Therefore, $2{\sin }^{-1}\left(\frac{5}{13}\right)={\cos }^{-1}\left(\frac{119}{169}\right)$.

Hence proved.