How do you prove that $e^{ia} + 1 = 2e^{i\frac{a}{2}}\cos(\frac{a}{2})$?

Question

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?

Answer 1

Hint:

$$e^{ia}+1=e^{\frac{ia}{2}}(e^{\frac{ia}{2}}+e^{-\frac{ia}{2}})$$

It might be useful to note that

$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$

Answer 2

hint

$$e^{ia}+1=1+\cos(a)\color{red}{+}i\sin(a)$$

$$=2\cos^2(\frac a2)\color{red}{+}2i\sin(\frac a2)\cos(\frac a2)$$

$$=2\cos(\frac a2)\Bigl(....\Bigr)$$

You need these identities $$1+\cos(X)=2\cos^2(\frac X2)$$ $$\sin(X)=2\sin(\frac X2)\cos(\frac X2)$$ and $$e^{iX}=\cos(X)+i\sin(X)$$

Answer 3

You could continue your way like this: $$=e^{ia} + \cos(a) + 2\sin^2(\frac{a}{2})=e^{ia} + \cos(a) + 2 - 2\cos^2(\frac{a}{2})=$$ $$= e^{ia} + \cos^2(\frac{a}{2})-\sin^2(\frac{a}{2}) + 2 - 2\cos^2(\frac{a}{2})=$$ $$=e^{ia} - \cos^2(\frac{a}{2})-\sin^2(\frac{a}{2}) + 2= e^{ia} - 1 + 2= e^{ia} + 1$$