Prove the included formula relating cos(nx) and cos(x) I'm struggling with the below problem. Can anyone shed some light on it?
Show that the below formula is a correct relation between $y = \cos n\theta$ and $x = cos \theta$ for all $n$:
$$ x = \frac 12 \sqrt[n]{y + \sqrt{y^2 - 1}} + \frac 12 \sqrt[n]{y - \sqrt{y^2 - 1}} $$
 A: This may not be exactly the proof you want, but here is a proof (here, we will only use the fact that $i=\sqrt{-1}$ a couple of times, and nothing else about the complex number $i$...I will make note of when we use this).:
We proceed first to prove the identity
$\cos(n\theta) + i\sin(n\theta) = (\cos(\theta) + i\sin(\theta))^n$
Clearly, this is true for $n=1$, so we proceed by induction.  Assume this is true up to $n=k$, and then prove it for $k+1$.
$(\cos(\theta) + i\sin(\theta))^{k+1} = (\cos(\theta)+i\sin(\theta))^k (\cos(\theta) + i\sin(\theta))$
Using our induction hypothesis,
$= (\cos(k\theta) + i\sin(k\theta))(\cos(\theta) + i\sin(\theta))$
$= (\cos(k\theta)\cos(\theta) - \sin(k\theta)\sin(\theta)) + i(\cos(k\theta)\sin(\theta) + \sin(k\theta)\cos(\theta))$
$= \cos((k+1)\theta) + i\sin((k+1)\theta)$
We have thus proven it for $n=k+1$, hence it is proven for all positive integers.
For negative integers, note that the formula is true for $n=0$, because in both cases, it is just 1.  Then,
$(\cos(\theta) + i\sin(\theta))^{-n} = ((\cos(\theta) + i\sin(\theta))^n)^{-1} = (\cos(n\theta) + i\sin(n\theta))^{-1}$
$ = (\cos(n\theta) - i\sin(n\theta)) = (\cos(-n\theta) + i\sin(-n\theta))$
The last two equalities follow by multiplying and dividing the last thing on the previous line by the complex conjugate.  Now, return to the formula we just proved
$\cos(n\theta) + i\sin(n\theta) = (\cos(\theta) + i\sin(\theta))^n$
Using this, we get
$(\cos(\theta) + i\sin(\theta)) = \sqrt[n]{\cos(n\theta) + i\sin(n\theta)}$
$= \sqrt[n]{\cos(n\theta) + \sqrt{-\sin^2(n\theta)}} = \sqrt[n]{\cos(n\theta) + \sqrt{-(1-\cos^2(n\theta))}} = \sqrt[n]{\cos(n\theta) + \sqrt{\cos^2(n\theta) - 1}}$
(In the second to last equality, we used $i=\sqrt{-1}$.)  Also, using the identity with $n$ replaced by $-n$, we get
$(\cos(\theta) - i\sin(\theta)) = \sqrt[n]{\cos(n\theta) - i\sin(n\theta)} = \sqrt[n]{\cos(n\theta) - \sqrt{-\sin^2(n\theta)}}$
$= \sqrt[n]{\cos(n\theta) - \sqrt{-(1-\cos^2(n\theta))}} = \sqrt[n]{\cos(n\theta) -\sqrt{\cos^2(n\theta) - 1}}$
(In the second to last equality, we used $i=\sqrt{-1}$.)  Add $1/2$ the first to $1/2$ the second to get
$\frac{1}{2}(\cos(\theta) + i\sin(\theta)) + \frac{1}{2}(\cos(\theta) - i\sin(\theta)) = \cos(\theta)$
$ = \frac{1}{2}\sqrt[n]{\cos(n\theta) + \sqrt{\cos^2(n\theta) - 1}} + \frac{1}{2}\sqrt[n]{\cos(n\theta) -\sqrt{\cos^2(n\theta) - 1}}$
Substituting $x = \cos(\theta)$, $y = \cos(n\theta)$ gives
$x = \frac{1}{2}\sqrt[n]{y + \sqrt{y^2-1}} + \frac{1}{2}\sqrt[n]{y-\sqrt{y^2-1}}$


Why the downvote?  If I made a mistake, point it out in comments and I will fix it.
