iarc6io

2022-07-19

How do you find the exact value of $\mathrm{arccos}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)$?

eishale2n

knowing that $\mathrm{sin}\left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}$
$\mathrm{arccos}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)=\mathrm{arccos}\left(\frac{\sqrt{3}}{2}\right)$
we know that $\mathrm{cos}\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}$
so, $\frac{\pi }{6}=\mathrm{arccos}\left(\frac{\sqrt{3}}{2}\right)$
$\mathrm{arccos}\left(\mathrm{sin}\left(\frac{\pi }{3}\right)\right)=\mathrm{arccos}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi }{6}$

valtricotinevh

By definition, $\mathrm{cos}\left(\frac{1}{2}\pi -\theta \right)=\mathrm{sin}\theta forall\theta$
$\therefore \mathrm{arccos}\left(\mathrm{sin}\left(\frac{1}{3}\pi \right)\right)=\mathrm{arccos}\left(\mathrm{cos}\left(\frac{1}{2}\pi -\frac{1}{3}\pi \right)\right)=\mathrm{arccos}\left(\mathrm{cos}\left(\frac{1}{6}\pi \right)\right)$
$=\frac{1}{6}\pi$

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