Question

1)sin6,3π = ??
2) cos 27π/8 = ??​

Answer 1

$1)Sin6,3\pi=Sin(6\pi+\frac{3\pi }{10})=Sin\frac{3\pi }{10}=Sin(\frac{\pi }{2}-\frac{\pi }{5})=\boxed{Cos\frac{\pi }{5}}\\\\\\2)Cos\frac{27\pi }{8}=Cos(2\pi +\frac{11\pi }{8})=Cos\frac{11\pi }{8}=Cos(\frac{3\pi }{2}- \frac{\pi }{8})=\boxed{-Sin\frac{\pi }{8}}$