How do you prove $e^{ia} + 1 = 2e^{i\frac{a}{2}}\cos(\frac{a}{2})$ where a is a real number?
I attempted to solve it, by using the second side and I couldn't find a correct result, sadly. I'm not even sure if my attempt is correct.
My solution:
$2cos^2(\frac{a}{2}) + 2isin(\frac{a}{2})cos(\frac{a}{2})$
By using cos and sin identities:
$2cos^2(\frac{a}{2}) + isin(a)$
$2cos(a) + 2sin^2(\frac{a}{2}) + isin(a)$
$e^{ia} + cos(a) + 2sin^2(\frac{a}{2})$
And that's it. The other further attempts have gone nowhere. Like I tried to use (cos^2 + sin^2 = 1) but it resulted in going nowhere, too. I just have no idea how to get 1 out.
Any ideas?
EDIT: I kept trying more and I think I figured it out.
By using cos and sin identities:
$cos(a) = cos^2(\frac{a}{2}) – sin^2(\frac{a}{2})$
By placing it in the equation I found earlier:
Then we get this
$e^{ia} + cos^2(\frac{a}{2}) + sin^2(\frac{a}{2})$
Which is
$e^{ia} + 1$
Is this correct?
Best Answer
Hint:
$$e^{ia}+1=e^{\frac{ia}{2}}(e^{\frac{ia}{2}}+e^{-\frac{ia}{2}})$$
It might be useful to note that