ВЫЧИСЛИТЬ cos 105°- cos 15°

Решите задачу:

$1)\; \; cos105^\circ -cos15^\circ =cos(90^\circ +15^\circ )-cos15^\circ =-sin15^\circ -cos15^\circ =\\\\=- \frac{\sqrt2(\sqrt3-1)}{4}-\frac{\sqrt2(\sqrt3+1)}{4}=-\frac{\sqrt2(\sqrt3-1+\sqrt3+1)}{4}=- \frac{\sqrt2\cdot 2\sqrt3}{4}=-\frac{\sqrt6}{2}$

$\star \; sin15^\circ =sin(45^\circ-30^\circ)=sin45^\circ\cdot cos30^\circ -cos45^\circ\cdot sin30^\circ =\\\\=\frac{\sqrt2}{2}\cdot \frac{\sqrt3}{2}-\frac{\sqrt2}{2}\cdot \frac{1}{2}=\frac{\sqrt2\cdot (\sqrt3-1)}{4} \\\\\star \; cos15^\circ =cos(45^\circ -30^\circ)= cos45^\circ \cdot cos30^\circ +sin45^\circ \cdot sin30^\circ =\\\\=\frac{\sqrt2}{2}\cdot \frac{\sqrt3}{2}+\frac{\sqrt2}{2}\cdot \frac{1}{2}= \frac{\sqrt2(\sqrt3+1)}{4}$

$2)\; \; cos105^\circ -cos15^\circ =-2\, sin \frac{105^\circ+15^\circ}{2}\cdot sin \frac{105^\circ -15^\circ}{2}=\\\\=-2\, sin60^\circ \cdot sin45^\circ =-2\cdot \frac{\sqrt3}{2}\cdot \frac{\sqrt2}{2}=- \frac{\sqrt3\cdot \sqrt2}{2}=-\frac{\sqrt6}{2}$