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Using trigonometry, one can easily establish that

$$\cos (\omega_1 t) + \cos (\omega_2 t) = 2 \cos \left[(\omega_1 - \omega_2) t/2\right] \cos \left[(\omega_1 + \omega_2) t/2\right]$$

But can we exploit $$\exp(i \theta) = \cos \theta + i \sin \theta$$ to arrive at the same answer using complex exponential form, i.e. starting from:

$$ \exp(i\omega_1 t) + \exp(i\omega_2 t) $$

and WITHOUT USING ANY TRIGONOMETRIC FORMULA?

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    You need to edit the first statement which is incorrect. There should be a factor of $\frac 12$ in the arguments – 2017-02-05
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    http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html – 2017-02-05
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    @DavidQuinn - Thanks for pointing out the mistake. Just made the edit. – 2017-02-05

2 Answers 2

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HINT:

$$e^{2iA}+e^{2iB}=e^{i(A+B))}\left(e^{i(A-B))}+e^{-i(A-B))}\right)$$

Now use Euler identity $e^{ix}=\cos x+i\sin x$ and subsequently, $$e^{i(A-B))}+e^{-i(A-B))}=2\cos(A-B)$$

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    Thanks a lot. One of those things that just don't occur to you at times! – 2017-02-05
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Of course you can:

$$ \cos (\omega_i t ) = \frac{e^{ i \omega_i t} + e^{-i \omega_i t}}{2} $$

Therefore:

$$ \begin{aligned} \cos (\omega_1 t ) + \cos (\omega_2 t) &= \frac{e^{ i \omega_1 t} + e^{-i \omega_1 t}}{2} + \frac{e^{ i \omega_2 t} + e^{-i \omega_2 t}}{2} \\ &= \frac{1}{2} \left [ e^{ i\frac{\omega_1 + \omega_2}{2} t} \left ( e^{ i\frac{\omega_1 - \omega_2}{2} t} + e^{ -i\frac{ \omega_1 - \omega_2 }{2} t } \right ) + e^{-i \frac{\omega_1 + \omega_2}{2} t} \left ( e^{ i\frac{\omega_1 - \omega_2}{2} t} + e^{ -i\frac{ \omega_1 - \omega_2 }{2} t } \right ) \right ] \\ &= \frac{1}{2} \left [ \left ( e^{ i\frac{\omega_1 + \omega_2}{2} t} + e^{ -i\frac{ \omega_1 + \omega_2 }{2} t } \right ) \cdot \left ( e^{ i\frac{\omega_1 - \omega_2}{2} t} + e^{ -i\frac{ \omega_1 - \omega_2 }{2} t } \right ) \right ] \\ &= 2 \cos \left ( \frac{\omega_1 + \omega_2}{2} t \right ) \cos \left ( \frac{\omega_1 - \omega_2}{2} t \right )\end{aligned} $$