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You know how $\sin(\theta) \cdot i + \cos \theta$ is in exponential form $e^{i\theta}$, why exactly is $\sin(\theta) \cdot i + \cos \theta = e^{i\theta}$?

Martin Sleziak
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M. C.
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    Are you asking for a proof of Euler's formula? – clathratus Jan 25 '19 at 00:31
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    Have you done any research at all on how this is shown? Wikipedia, multiple questions here, any complex analysis book, most calculus books..... Please clarify what it is that you aren't sure about, rather than just asking for a rehashing of something that's quite available. –  Jan 25 '19 at 00:44
  • If anyone can delete this question, that would be great! It is a duplicate. – M. C. Jan 28 '19 at 21:36

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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

James
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