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We're having trouble proving this trig identity:

$$e^{i \alpha} + e^{i \beta} = 2e^{\frac{i(\alpha + \beta)}{2}}\cos(\frac{\alpha-\beta}{2})$$


We've tried various manipulations using basic trig identities (too lengthy to show here) and come up with nasty equations that don't lead anywhere.

Can you point us in the right direction?

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  • $\begingroup$ I assume $\alpha, \beta$ are real. $\endgroup$ – N. S. Nov 7 '11 at 3:34
  • $\begingroup$ @N.S. yes, they are $\endgroup$ – smackcrane Nov 7 '11 at 3:35
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You need to show that

$$\cos(\alpha)+\cos(\beta)+i[\sin(\alpha)+\sin(\beta)]= 2 [\cos(\frac{\alpha+\beta}{2})+i \sin (\frac{\alpha+\beta}{2})]\cos(\frac{\alpha-\beta}{2}) \,.$$

Identify the real and imaginary parts, and use the formulas for $\cos(x)\cos(y)$ and $\sin(x) \cos(y)$ respectively...

Alternately, you can use that $\alpha=\frac{\alpha+\beta}{2}+\frac{\alpha-\beta}{2}$ and $\beta=\frac{\alpha+\beta}{2}-\frac{\alpha-\beta}{2}$.

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If you use the fact that $\cos((\alpha-\beta)/2)=\frac{\exp(i(\alpha-\beta)/2)+\exp(-i(\alpha-\beta)/2)}{2}$ lots of things cancel.

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Just work on the right side, it is easy to show using definition of cosine over the complex numbers:

$$cos(z)=\frac {e^{iz}+e^{-iz}}{2}$$

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