<s>[INST] 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 $99\times \frac{5\pi}{6}$?
 [/INST] $\cos \left(99 \times \frac{5 \pi}{6} \right) = \cos \left(33 \times \frac{5 \pi}{2} \right) = \cos \left(\frac{165 \pi}{2} \right) = 0$ since $\cos(\frac{n \pi}{2}) = 0$ whenever $n$ is odd.
$\sin \left(99 \times \frac{5 \pi}{6} \right) = \sin \left(33 \times \frac{5 \pi}{2} \right) = \sin \left(\frac{165 \pi}{2} \right) = \sin \left(82 \pi + \frac{\pi}{2} \right) = 1$ since $\sin(2n \pi + \frac{\pi}{2}) = 1$.
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