Question

Помогите пожалуйста математика

Answer 1

Ответ:

Ответ на фото

Пошаговое объяснение:

Answer 2

Ответ:

1.

$30^\circ =  \frac{\pi}{6}\\\\60^\circ = \frac{\pi}{3}$

2.

$120^\circ = \frac{2\pi}{3}\\\\135^\circ = \frac{3\pi}{4}$

3.

$10^\circ = \frac{\pi}{18}\\\\36^\circ = \frac{\pi}{5}$

4.

$720^\circ = 4\pi\\\\450^\circ = \frac{5\pi}{2}$

5.

$\frac{\pi}{4} = \frac{180^\circ}{4} = 45^\circ\\\\\frac{2\pi}{3} = \frac{2*180^\circ}{3} = 120^\circ$

6.

$\frac{5\pi}{3} = \frac{5*180^\circ}{3} = 300^\circ\\\\\frac{11\pi}{6} = \frac{11*180}{6} = 330^\circ$

7.

$\frac{\pi}{12} = \frac{180^\circ}{12} = 15^\circ\\\\\frac{7\pi}{10} = \frac{7*180^\circ}{10} = 126^\circ$

8.

$3\pi = 3*180^\circ = 540^\circ\\\\2,5\pi = \frac{5\pi}{2} = \frac{5*180^\circ}{2} = 450^\circ$

9.

$sin90^\circ = 1\\\\cos180^\circ = -1$

10.

$tg360^\circ = 0\\\\ctg270^\circ = 0$

11.

$sin(-\frac{\pi}{2}) = sin(-90^\circ) = -sin(90^\circ) = -1\\\\cos(-\pi) = cos(-180^\circ) = cos(180^\circ) = -1$

12.

$sin(-45^\circ) = -sin(45^\circ) = -\frac{\sqrt{2}}{2}\\\\cos(-60^\circ) = cos(60^\circ) = \frac{1}{2}$

13.

$tg(-30^\circ) = -tg(30^\circ) = -\frac{\sqrt{3}}{3}\\\\ctg(-30^\circ) = -ctg(30^\circ) = -\sqrt{3}$

Удачи!