Вычислить: (4 cos (25pi /12) + 4 sin (19pi/12))/ (-4 cos (pi/12) - 2 cos (11pi/12))

Решите задачу:

$\frac{4cos\frac{25\pi}{12}+4sin\frac{19\pi }{12}}{-4cos\frac{\pi}{12}-2cos\frac{11\pi}{12}} = \frac{4cos(2\pi +\frac{\pi}{12})+4sin(2\pi -\frac{5\pi}{12})}{-4cos\frac{\pi}{12}-2cos(\pi -\frac{\pi}{12})} = \frac{4(cos\frac{\pi}{12}-sin\frac{5\pi}{12})}{-2(2cos\frac{\pi}{12}+c os\frac{\pi}{12})}=\\\\=-2\cdot \frac{cos\frac{\pi}{12}+cos\frac{\pi}{12}}{3cos\frac{\pi}{12}}=-2\cdot \frac{2cos\frac{\pi}{12}}{3cos\frac{\pi}{12}} =-\frac{4}{3}$

$\star \; \; \; \; \; sinx=cos(\frac{\pi}{2}-x)\; \; \Rightarrow \\\\sin\frac{5\pi}{12}=cos(\frac{\pi}{2}-\frac{5\pi}{12})=cos(\frac{6\pi}{12}-\frac{5\pi}{12})=cos\frac{\pi}{12}$