Question

Помогите пожалуйста arcsin(-√3:2) -2arccos(-√3:2) -arcctg0

Answer 1

$\displaystyle arcsin( - x) = - x; \: - \frac{\pi}{2} \leqslant x \leqslant \frac{\pi}{2}$

$\displaystyle arccos ( - x) = \pi - arccos x; \: 0 \leqslant x \leqslant \pi$

$\displaystyle arcctg x = x; \: 0 \leqslant x \leqslant \pi$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀Решение

$\displaystyle arcsin \bigg( - \frac{ \sqrt{3} }{2} \bigg) - 2arccos \bigg( - \frac{ \sqrt{3} }{2} \bigg) - arcctg0 =$

$\displaystyle - \frac{\pi}{3} - 2 \: * \: \bigg(\stackrel{6/}{}\pi - \frac{\pi}{6} \bigg) - \frac{\pi}{2} = - \frac{\pi}{3} -$

$\displaystyle - \not2\: * \: \frac{5\pi}{ \not6} - \frac{\pi}{2} = - \frac{\stackrel{2/}{}\pi}{3} - \frac{\stackrel{2/}{}5\pi}{3} - \frac{\stackrel{3/}{}\pi}{2} =$

$\displaystyle = \frac{ - 2\pi - 10\pi - 3\pi}{6} = - \frac{\stackrel{5/}{} \not15\pi}{\stackrel{2/}{}6} = \bf2,5\pi$