Question-140192

Question Number 140192 by meetbhavsar25 last updated on 05/May/21

Answered by mr W last updated on 05/May/21

s a y \tan^{- 1} x = t

x = \tan t

\cos t = \tan t = \frac{\sin t}{\cos t}

\cos^{2} t = \sin t

\sin^{2} t + \sin t - 1 = 0

\sin t = \frac{- 1 + \sqrt{5}}{2}

\frac{x^{2}}{2} = \frac{\tan^{2} t}{2} = \frac{\sin^{2} t}{{2 \cos}^{2} t} = \frac{\sin t}{2} = \frac{\sqrt{5} - 1}{4}

\Rightarrow \cos^{- 1} (\frac{x^{2}}{2}) = \cos^{- 1} (\frac{\sqrt{5} - 1}{4}) = \frac{2 π}{5} = 72 °

Commented by mr W last updated on 05/May/21

\cos (\frac{x^{2}}{2}) {d o e s n}^{'} t m a k e m u c h s e n s e,

s o i c h a n g e d t h e q u e s t i o n t o \cos^{- 1} (\frac{x^{2}}{2}) .

Commented by meetbhavsar25 last updated on 05/May/21

Thank you and you were correct....i mistakenly wrote cos instead of cos-¹