Proving something about complex

**dnlstffrd** · August 28th 2007, 01:06 PM

I need to prove that e^(it) + e^(is) = 2cos((t-s)/2)e^(i(s+t)/2). I have used lns and gotten it to -ts = i((s+t)/2)ln(2cos((t-s)/2), but I don't know where to go from there.

**Jhevon** · August 28th 2007, 01:25 PM

Originally Posted by dnlstffrd

I need to prove that e^(it) + e^(is) = 2cos((t-s)/2)e^(i(s+t)/2). I have used lns and gotten it to -ts = i((s+t)/2)ln(2cos((t-s)/2), but I don't know where to go from there.

It's not that bad. Here's one way to do it. You need to know three things:

Euler's Formula:

$e^{i \theta} = \cos \theta + i \sin \theta$

Product-to-sum formulas for sine and cosine:

$\sin \alpha + \sin \beta = 2 \sin \frac {\alpha + \beta}{2} \cos \frac {\alpha - \beta}{2}$

and finally,

$\cos \alpha + \cos \beta = 2 \cos \frac {\alpha + \beta}{2} \cos \frac {\alpha - \beta}{2}$

Now, let's get to it:

${\color {red}e^{it}} + {\color {blue} e^{is}} = { \color {red} \cos t + i \sin t} + { \color {blue} \cos s + i \sin s}$ ...........................applied Euler's formula

$= \left( \cos t + \cos s \right) + i \left( \sin t + \sin s \right)$ ....................................rearranged the terms

$= \left( 2 \cos \frac {s + t}{2} \cos \frac {t - s}{2} \right) + i \left( 2 \sin \frac {s + t}{2} \cos \frac {t - s}{2} \right)$ .........applied the sum-to-product formulas

$= 2 \cos \frac {t - s}{2} \left[ { \color {red} \cos \frac {s + t}{2} + i \sin \frac {s + t}{2}} \right]$ ...............................factored out the common $2 \cos \frac {t - s}{2}$

$= 2 \cos \frac {t - s}{2}e^{i \frac {s + t}{2}}$

as desired

**dnlstffrd** · August 28th 2007, 01:38 PM

I totally forgot about Euler's formula, once I remembered that it was a cinch lol

**ThePerfectHacker** · August 28th 2007, 01:43 PM

I have an geometric approach to this. Try to see if you can come up with it.

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