Предмет: Алгебра
Автор: jeremiahvaleska
Вопрос задан 2 года назад

1) cos60° – sin60° + ctg60° – tg60°;
2) -sin30° + cos30° – ctg30° + tg30°;
3) cos45° – tg45° – sin45° + ctg45°;
4) sin0° – cos30° + tg45° – ctg60°;
5) -cos0° + tg30° - ctg45° + sin60°;
6) tg0° – ctg90° - sin0° – cos90°.

Приложения:

Ответы

Ответ дал: Miroslava227

Ответ:

$\cos(60^{\circ} ) - \sin(60^{\circ} ) + ctg(60^{\circ} ) - tg(60^{\circ} ) = \\ = \frac{1}{2} - \frac{ \sqrt{3} }{2} + \frac{ \sqrt{3} }{3} - \sqrt{3} = \\ = \frac{3 - 3 \sqrt{3} + 2 \sqrt{3} - 6 \sqrt{3} }{6} = \\ = \frac{3 - 7 \sqrt{3} }{6}$

$- \sin(30^{\circ} ) + \cos(30^{\circ} ) - ctg(30^{\circ} ) + tg(30^{\circ} ) = \\ = - \frac{1}{2} + \frac{ \sqrt{3} }{2} - \sqrt{3} + \frac{ \sqrt{3} }{3} = \\ = \frac{ - 3 + 3 \sqrt{3} - 6 \sqrt{3} + 2 \sqrt{3} }{6} = \\ = \frac{ - 3 - \sqrt{3} }{6} = - \frac{3 + \sqrt{3} }{6}$

$\cos(45^{\circ} ) - tg(45^{\circ} ) - \sin(45^{\circ} ) + ctg(45^{\circ} ) = \\ = \frac{ \sqrt{2} }{2} - 1 - \frac{ \sqrt{2} }{2} + 1 = 0$

$\sin(0^{\circ} ) - \cos(30^{\circ} ) + tg(45^{\circ} ) - ctg(60^{\circ} ) = \\ = 0 - \frac{ \sqrt{3} }{2} + 1 - \frac{ \sqrt{3} }{3} = \frac{ - 3 \sqrt{3} - 2 \sqrt{3} }{6} = - \frac{5 \sqrt{3} }{6}$

$- \cos(0^{\circ} ) + tg(30^{\circ} ) - ctg(45^{\circ} ) + \sin(60^{\circ} ) = \\ = - 1 + \frac{ \sqrt{3} }{3} - 1 + \frac{ \sqrt{3} }{2} = \\ = \frac{ - 6 + 2 \sqrt{3} + 3 \sqrt{3} }{6} = \frac{5 \sqrt{3} - 6 }{6}$