Question

Solve ${\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$

Answer 1

${\sin }^{-1}\left(\frac{5x+12\sqrt{1-x^2}}{13}\right)$

putting $x=\cos \theta ,\theta ={\cos }^{-1}x$

$={\sin }^{-1}\left(\frac{5\cos \theta +12\sqrt{1-{\cos }^2\theta }}{13}\right)$

$={\sin }^{-1}\left(\frac{5\cos \theta +12\sqrt{{\sin }^2\theta }}{13}\right)$

$={\sin }^{-1}\left(\frac{5\cos \theta +12\sin \theta }{13}\right)$

$={\sin }^{-1}\left(\frac{5\cos \theta }{13}+\frac{12\sin \theta }{13}\right)$

putting $sin\varphi =\frac{5}{13}$ and $cos\varphi =\frac{12}{13}$

$tan\theta =\frac{5}{12}$

$={\sin }^{-1}\left(\sin \varphi \cos \theta +\cos \varphi \sin \theta \right)$

$={\sin }^{-1}\left(\sin \left(\varphi +\theta \right)\right)=\varphi +\theta$

$={\tan }^{-1}\left(\frac{5}{12}\right)+{\cos }^{-1}x$