Question

Ребята, объясните мне пожалуйста по какому принципу нужно решать это... (10 класс) ​

Answer 1

$1)Sin(\frac{13\pi }{6})=Sin(2\pi+\frac{\pi }{6})=Sin\frac{\pi }{6}=\frac{1}{2}\\\\Cos(\frac{13\pi }{6})=Cos(2\pi+\frac{\pi }{6})=Cos\frac{\pi }{6}=\frac{\sqrt{3}}{2} \\\\2)Sin(-\frac{8\pi }{3})=-Sin(2\pi +\frac{2\pi }{3})=-Sin\frac{2\pi }{3}=-Sin(\pi-\frac{\pi }{3})=-Sin\frac{\pi }{3}=-\frac{\sqrt{3}}{2}\\\\Cos(-\frac{8\pi }{3})=Cos(2\pi +\frac{2\pi }{3})=Cos\frac{2\pi }{3}=Cos(\pi-\frac{\pi }{3})=-Cos\frac{\pi }{3}=-\frac{1}{2}$

$3)Sin(\frac{23\pi }{6})=Sin(4\pi-\frac{\pi }{6})=Sin(-\frac{\pi }{6})=-Sin\frac{\pi }{6}=-\frac{1}{2}\\\\Cos(\frac{23\pi }{6})=Cos(4\pi-\frac{\pi }{6})=Cos(-\frac{\pi }{6})=Cos\frac{\pi }{6}=\frac{\sqrt{3}}{2}\\\\4)Sin(-\frac{11\pi }{3})=-Sin(\frac{11\pi }{3})=-Sin(4\pi-\frac{\pi }{3})=-Sin(-\frac{\pi }{3})=Sin\frac{\pi }{3}=\frac{\sqrt{3}}{2}\\\\Cos(-\frac{11\pi }{3})=Cos(\frac{11\pi }{3})=Cos(4\pi-\frac{\pi }{3})=Cos(-\frac{\pi }{3})=Cos\frac{\pi }{3}=\frac{1}{2}$