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Answer
If you remember series, notice that
$$ e^{i x } = \sum_{n \geq 0} \frac{ i^n x^n }{n!} $$
Now, notice that $i^2 = -1 $ but $i^{3} = -i$, and $i^4 = 1 $ and $i^5 = i$ so on, and since
$$ \sin x = \sum_{n \geq 0} \frac{ (-1)^n x^{2n+1 } }{(2n+1)!} \; \; \text{and} \; \;\cos x = \sum_{n \geq 0} \frac{ (-1)^n x^{2n } }{(2n)!} $$
after breaking the $n$ in the first summation for even cases and odd cases and seeing in the third line how the $i's$ alternate, one obtains the result
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