# 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 $99\times \frac{5\pi}{6}$?

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## 3 Answers

Well, $99\cdot \frac{5\pi}{6}=33\cdot \frac{5\pi}{2}=\frac{165\pi}{2}$, so

$\sin\left(\frac{165\pi}{2}\right)=\sin\left(82\pi + \frac{\pi}{2}\right)=\sin(\frac{\pi}{2})=1$. Similarly, $\cos\left(\frac{165\pi}{2}\right)=0$.

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

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$\cos \left(99 \times \frac{5 \pi}{6} \right) = \cos \left(33 \times \frac{5 \pi}{2} \right) = \cos \left(\frac{165 \pi}{2} \right) = 0$ 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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