Вычислите cos 105 градусов, sin 300 градусов и tg 41Пи/4

Решите задачу:

$sin300^\circ =sin(360^\circ -60^\circ )=-sin60^\circ =-\frac{\sqrt3}{2}\\\\tg\frac{41\pi}{4}=tg(10\pi +\frac{\pi}{4})=tg\frac{\pi}{4}=1\\\\cos105^\circ =cos(90^\circ +15^\circ)=-sin15^\circ=-sin(45^\circ-30^\circ)=\\\\=-sin45\cdot cos30+cos45\cdot sin30=-\frac{\sqrt2}{2}\cdot \frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot \frac{1}{2}=\\\\=\frac{\sqrt2(1-\sqrt3)}{4}=\frac{\sqrt2-\sqrt6}{4}$