Trigonometric functions, arcus functions

Quote:

Originally Posted by sf1903

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 =?

$\sin A = \cos (\frac{\pi}{2} - A)$ and

$\sin (2\pi+A) =\sin A$

$\cos ^{-1} (\cos A) = A$

ok

$\cos ^{-1} (\sin \frac{49\pi}{8})$ but

$\frac{49\pi}{8} = 6\pi + \frac{\pi}{8} = 3(2\pi) + \frac{\pi}{8}$

so

$\cos ^{-1} ( \sin \frac{\pi}{8})$

$\cos ^{-1} (\cos (\frac{\pi}{2} - \frac{\pi}{8}))=\frac{\pi}{2} - \frac{\pi}{8}$