Смотреть решения всех заданий с фото
Вопрос:

021.16. a) arccos (\(-\frac{1}{2}\)) + arcsin (\(-\frac{1}{2}\)); б) arccos (\(-\frac{\sqrt{2}}{2}\)) - arcsin (-1); в) arccos (\(-\frac{\sqrt{3}}{2}\)) + arcsin (\(-\frac{\sqrt{3}}{2}\)); г) arccos \(\frac{\sqrt{2}}{2}\) - arcsin \(\frac{\sqrt{3}}{2}\);

Ответ:

a) arccos (\(-\frac{1}{2}\)) + arcsin (\(-\frac{1}{2}\)) = \(\frac{2\pi}{3} + (-\frac{\pi}{6}) = \frac{4\pi}{6} - \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}\). б) arccos (\(-\frac{\sqrt{2}}{2}\)) - arcsin (-1) = \(\frac{3\pi}{4} - (-\frac{\pi}{2}) = \frac{3\pi}{4} + \frac{2\pi}{4} = \frac{5\pi}{4}\). в) arccos (\(-\frac{\sqrt{3}}{2}\)) + arcsin (\(-\frac{\sqrt{3}}{2}\)) = \(\frac{5\pi}{6} + (-\frac{\pi}{3}) = \frac{5\pi}{6} - \frac{2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}\). г) arccos \(\frac{\sqrt{2}}{2}\) - arcsin \(\frac{\sqrt{3}}{2}\) = \(\frac{\pi}{4} - \frac{\pi}{3} = \frac{3\pi}{12} - \frac{4\pi}{12} = -\frac{\pi}{12}\).

Похожие