How to get from $2^{99} \cdot (\cos{(99\times \frac{5\pi}{6}) + i\cdot \sin{(99\times \frac{5\pi}{6})}})$ to $0+2^{99}i$? 
How do I get from 2nd last to last step? How did they simplify cos & sin $99\times \frac{5\pi}{6}$?

$99\cdot \frac {5 \pi}{6}=82.5\pi $ and you can ignore the multiples of $2 \pi$
