Question

12 arcsin √2/2+4 arccos 1 - 16 arctg 1 - 24 arcctg √3

Answer 1

Ответ:

$12arcsin( \frac{ \sqrt{2} }{2} ) + 4arccos(1) - 16arctg(1) - 24arcctg( \sqrt{3} ) = \\ = 12 \times \frac{\pi}{4} + 0 - 16 \times \frac{\pi}{4} - 24 \times \frac{\pi}{6} = \\ = 3\pi - 4\pi - 6\pi = - 7\pi$