Question

Помогите, срочно! Даю 50 баллов!

Answer 1

Ответ:

$1-\frac{\sqrt{3}}{2}$

Объяснение:

$\sqrt{2} cos(-1755^\circ+360^\circ*5)-\sqrt{3} sin(-930^\circ+360^\circ*3)+\\ \\ +\frac{\sqrt{3} }{3}tg(-600^\circ+360^\circ*2) +ctg(-1395^\circ+360^\circ*4)= \\ \\ \sqrt{2} cos45^\circ-\sqrt{3} sin150^\circ+\frac{\sqrt{3} }{3}tg120^\circ +ctg45^\circ=\sqrt{2}*\frac{\sqrt{2}}{2} -\sqrt{3}*\frac{1}{2} + \\ \\ +\frac{\sqrt{3}}{3} *(-\sqrt{3})+1=1-\frac{\sqrt{3}}{2} -1+1=1-\frac{\sqrt{3}}{2}$

Answer 2

$1) cos(-1775^{\circ}) = cos(-8\pi -315^{\circ}) = cos(-315^{\circ}) = cos(45^{\circ}) = \frac{\sqrt{2}}{2}$

$2) \sqrt{2} \cdot \frac{\sqrt{2}}{2} = \frac{2}{2} = 1$

$3) sin(-930^{\circ}) = -sin(30^{\circ}+5\cdot180^{\circ}) = -sin(-30^{\circ}) = sin 30^{\circ} = 0,5$

$4) \sqrt{{3}} \cdot sin30^{\circ}} = \frac{\sqrt{3}}{2}$

$5) tg(-600^{\circ}) = -\sqrt{3}$

$6) \frac{-\sqrt{3}\cdot\sqrt{3}}{3} = -\frac{3}{3} = -1$

$7) ctg (-1395^{\circ}) = 1$

Ответ: $1 - \frac{\sqrt{3}}{2}} - 1 + 1 = 1 - \frac{\sqrt{3}}{2}}$