cos-1-sin-41-sin-19-3-4-

Question Number 149259 by john_santu last updated on 04/Aug/21

\cos^{- 1} (\sqrt{\sin (41 °) \sin (19 °) + \frac{3}{4}}) = ?

Answered by john_santu last updated on 04/Aug/21

\Leftrightarrow \sin 41 ° \sin 19 ° = \frac{- 2 \sin 4 l ° \sin 19 °}{- 2}

= \frac{\cos 60 ° - \cos 22 °}{- 2} = \frac{\cos 22 ° - \frac{1}{2}}{2}

= \frac{1}{2} \cos 22 ° - \frac{1}{4}

\Rightarrow \cos^{- 1} (\sqrt{\frac{1}{2} \cos 22 ° + \frac{1}{2}}) =

\cos^{- 1} (\sqrt{\frac{1}{2} (2 \cos^{2} 11 ° - 1 + 1)})

\cos^{- 1} (\sqrt{\cos^{2} 11 °}) = \cos^{- 1} (\cos 11 °)

= 11 °