Formula for Nested Radicals I know that: $$\sqrt{2+\sqrt{2+\sqrt{2+...\sqrt{2}\; (upto\; n\; times)}}}=2\cos(2^{-n-1}\:\pi)$$
I was wondering whether such a formula exists for $$\sqrt{3+\sqrt{3+\sqrt{3+...\sqrt{3}\; (upto\; n\; times)}}}$$
or in general for,
$$\sqrt{k+\sqrt{k+\sqrt{k+...\sqrt{k}\; (upto\; n\; times)}}}$$
I've tried scaling the formula for 2 to get an approximate result. 
For example for the 6 case:
$$\sqrt{6+\sqrt{6+\sqrt{6+...\sqrt{6}\; (upto\; n\; times)}}}\approx\left(\frac{2\cos(2^{-n-1})-\sqrt2}{2-\sqrt2}\right)(3-\sqrt6)+\sqrt6$$

Let $$f(k,n)=\underbrace{\sqrt{k+\sqrt{k+\sqrt{\ldots+\sqrt k}}}}_n $$
We know that $\cos\frac x2=\pm\frac12\sqrt{1+\cos x}$. Therefore if $f(n,k)=a\cos b$ then $$f(n+1,k)=\sqrt{k+f(n,k)}=2\sqrt k\cdot\frac12\sqrt{1+\frac{f(n,k)}{k}}=2\sqrt k\cdot\frac12\sqrt{1+\frac ak\cos b}. $$
This works out nicely with our haf-angle formula only if $a=k$ and then becomes
$$ f(n+1,k)=2\sqrt k\cdot\frac12\sqrt{1+\cos b}=2\sqrt k\cos\frac b2.$$
We'd like to have $2\sqrt k=a=k$ again, but that works only if $k=2$.
In other words, there won't be formulas of the form $f(n,k)=a_k\cdot \cos 2^{-n}\alpha_k$ except for $k=2$.
