How to solve $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2} } }$ for $\cos(\alpha)$ using half-angle formula. I have $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2}  } }  $
How can I get $\cos(\alpha ) $ from this?
I know this identitiy. 
$\cos(\frac{\alpha }{2} )=\sqrt{\frac{1+\cos(\alpha )  }{2} }  $
But just cant figure out, how to do it.
 A: HINT:
You have $$\cos(\alpha/2)=\sqrt{\frac{1+\cos(\alpha)}{2}}=\frac{a}{\sqrt{a^2+b^2}}$$
Just square both sides and solve for $\cos(\alpha)$
A: $\cos(\frac{\alpha }{2} )=\frac{a}{\sqrt{a^{2}+b^{2}  } }  $
$\cos^2(\frac{\alpha }{2} )=\frac{a^2}{a^2+b^2}$
$\frac{1}{2}(1 + \cos(\alpha))=\frac{a^2}{a^2+b^2}$
$\cos(\alpha)=\frac{2a^2}{a^2+b^2} - 1$

half-angle formula
$\cos^2(\frac{\alpha }{2} ) = \frac{1}{2}(1 + \cos(\alpha))$
A: If we have that $a,b$ are given then we can use the above answers to say that 
$$\sqrt{\frac{1+\cos(\alpha)}{2}}=\frac{a}{\sqrt{a^2+b^2}}$$
Then 

$$\cos(\alpha) = \frac{2a^2}{a^2+b^2}-1$$

If instad you have $\sin(\alpha/2) = \frac{a}{\sqrt{a^2+b^2}}$ you can use that $\sin(\alpha/2) = \sqrt{1 - \cos^2(\alpha/2)} = \sqrt{1 - \frac{1+\cos(\alpha)}{2}} = \frac{a}{\sqrt{a^2+b^2}}$ and you solve again for $\cos(\alpha)$. 
If you want to exchange $\cos(\alpha)$ for $\sin(\alpha)$ it is always possible using $\cos^2+\sin^2=1$ and the identities that came from that. I think that in the comments was that what you pointed out. 
$\cos^2(2\alpha) = \cos^2(\alpha)-\sin^2(\alpha) = 2\cos^2(\alpha)-1$
$\sin^2(2\alpha) = 4\sin^2(\alpha)\cos^2(\alpha)=4\sin^2(\alpha)(1 - \sin^2(\alpha))$
So what possibly be other interpretation of your question is that given
$$\cos(\alpha/2) = \frac{a}{\sqrt{a^2+b^2}}$$
what is $\sin(\alpha)$? We know that, because of this we have that 
$$\cos(\alpha/2) = \frac{b}{\sqrt{a^2+b^2}}$$
So that $\cos^2+\sin^2 = 1$ and then we can use that 

$$\sin(\alpha) = 2\cos(\alpha/2)\sin(\alpha/2) = \frac{2ab}{a^2+b^2}$$

And the answers quoted respect the fundamental equation of trigonometry.
