Determine arccos(sin 49*pi /8)
The answer can be written in the form a/b where a/b is an abbreviated fraction.
a =?
b =?
$\displaystyle \sin A = \cos (\frac{\pi}{2} - A) $ and
$\displaystyle \sin (2\pi+A) =\sin A $
$\displaystyle \cos ^{-1} (\cos A) = A $
ok
$\displaystyle \cos ^{-1} (\sin \frac{49\pi}{8})$ but
$\displaystyle \frac{49\pi}{8} = 6\pi + \frac{\pi}{8} = 3(2\pi) + \frac{\pi}{8} $
so
$\displaystyle \cos ^{-1} ( \sin \frac{\pi}{8})$
$\displaystyle \cos ^{-1} (\cos (\frac{\pi}{2} - \frac{\pi}{8}))=\frac{\pi}{2} - \frac{\pi}{8}$