Question

Алгебра. Тригонометрия

Answer 1

a)

$\sin(\alpha) = \sqrt{1 - \cos^2(\alpha)} = -\frac{4}{5}\\\sin(\pi/3 - \alpha) = \sin(\pi/3)\cos(\alpha) - \cos(\pi/3)\sin(\alpha) = -\frac{3\sqrt{3}}{10} + \frac{2}{5}$

б)

$\cos(\beta) = \sqrt{1 - \sin^2(\beta)} = \frac{8}{17}\\\cos(\pi/6 + \beta) = \cos(\beta)\cos(\pi/6) - \sin(\beta)\sin(\pi/6) = \frac{4\sqrt{3}}{17} + \frac{15}{34}$

в)

$tg(\pi/4 + \alpha) = \frac{tg(\pi/4) + tg(\alpha)}{1 - tg(\pi/4)tg(\alpha)} = \frac{1 + tg(\alpha)}{1 - tg(\alpha)} = 2$

г)

Тут ошибка в условии, ведь $tg(\alpha) > 0, \pi < \alpha < 3\pi/2$.

д)

$\pi / 2 < \alpha, \beta < \pi$

$\cos(\alpha) = -0.8, \cos(\beta) = -0.6\\\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) = 0.28\\\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) = 0$