Question

Решить. ответ некопировать и не спамить!

Answer 1

$1)\; \; sin \frac{13\pi }{6}=sin(2\pi +\frac{\pi}{6})=sin\frac{\pi}{6}= \frac{1}{2}\\\\2)\; \; tg(- \frac{11\pi}{6})=-tg(2\pi -\frac{\pi}{6})=tg\frac{\pi}{6}=\frac{\sqrt3}{3}\\\\3)\; \; cos\pi +ctg\frac{4\pi}{3}=-1+ctg(\pi +\frac{\pi}{3})=-1+ctg\frac{\pi}{3}=-1+\frac{\sqrt3}{3}\\\\4)\; \; tg \frac{\pi }{4}\cdot ctg(-\frac{\pi }{4})+cos \frac{3\pi }{2}\cdot sin \frac{\pi }{2}=1\cdot (-1)+0\cdot 1=-1$