Simplifying imaginary term of jt in Fourier

I can't figure this out. Don't blame me, but please answer this question. I want to simplify this term:

$3(e^{5it} + e^{-5it})$

It would be nice to see a detailed workout. I know the answer is $2\sin{3t}$, but how to get to this is what I want to know.

Thanks.

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Hint: $\mathrm e^{\mathrm i t} \equiv \cos t + \mathrm i \sin t$.

P.S. The answer is wrong. See if you can correct it!

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HINT Two things you need to know:

First, use Euler's formula, $e^{it}=cos(t)+isin(t)$

$\therefore$ $$3*(e^{5it}+e^{-5it})$$ $$3*(cos(5t)+i*sin(5t))~+~(cos(-5t)+i*sin(-5t))$$

Second, since cosine is an even function, $cos(t)=cos(-t)$ And since sine is an odd function, $-sin(t)=sin(-t)$

Second hint, the answer is not what you claim it is.

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Use Euler's formula. Also, the answer is not $2\sin 3t$.

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