ОБЧИСЛИТИ(У ГРАДУСАХ) arccos(sin 40π \ 6) - 3 arctg( tg 40π \ 6 )

[latex]1)\; sin\frac{40\pi}{6}=sin\frac{20\pi}{3}=sin(6\pi +\frac{2\pi}{3})=sin\frac{2\pi}{3}=sin (\frac{\pi}{2}+\frac{\pi}{6})=cos\frac{\pi}{6}\\\\2)\; arccos(sin\frac{40\pi}{6})=arccos(cos\frac{\pi}{6})=arccos\frac{\sqrt3}{2}=\frac{\pi}{6}\\\\3)\; tg\frac{40\pi}{6}=tg(6\pi +\frac{2\pi}{3})=tg\frac{2\pi}{3}=tg(\pi -\frac{\pi}{3})=-tg\frac{\pi}{3}\\\\4)\; arctg(tg\frac{40\pi}{6})=arctg(-tg\frac{\pi}{3})=-arctg(tg\frac{\pi }{3})=-arctg(\sqrt3)=-\frac{\pi}{3}[/latex] [latex]5)\; \; arccos(sin\frac{40\pi}{6})-3arctg(tg\frac{40\pi}{6})=\frac{\pi}{6}-3\cdot (-\frac{\pi}{3})=\\\\=\frac{\pi}{6}+\pi =\frac{7\pi}{6}=7\cdot \frac{\pi}{6}=7\cdot 30^{\circ }=210^{\circ}[/latex]