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