From another math.stackexchange.com question we have
Can this expression be written as a nested square root only? ie,
where all of the a's are integers?
Edit: From the comment, it seems that it may be more interesting to look at
After doing some more research, I found that
This was based on this answer to a related question.. So it appears that it would be possible with an infinite nested square root. How about a finite nested square root? Would that be possible?
For all $A$ and $a_i$ it follows: $$A=\sqrt{1+a_1(A+1)+(A-a_1-1)\sqrt{1+a_2(A+2)+(A-a_2)\sqrt{1+a_3(A+3)+\cdots}}}$$
Let $$A=\cos \bigg(\frac{2\pi}{17}\bigg)$$ and let $$a_i=\frac{1}{A+i}$$ then, if $B_n=A-a_n+n-2$, we have
$$\cos\bigg(\frac{2\pi}{17}\bigg)=\sqrt{2+B_1\sqrt{2+B_2\sqrt{2+B_3\sqrt{2+\cdots}}}}$$
Any number cna be expressed as an infinite-nested radical, but given that $\cos(2\pi/17)$ is pretty obscure, I think this formula is the best to apply, since it works with infinite parameters and is not periodic (making it interesting!).
Here is an interesting one for cubed root (but periodic). Note that $$\color{red}{\cos\bigg(\frac{2\pi}{17}\bigg)}^3=\frac 14\cos \bigg(\frac{6\pi}{17}\bigg)+\frac 34\color{red}{\cos\bigg(\frac{2\pi}{17}\bigg)}$$
thus
$$\cos \frac{2\pi}{17}=\sqrt[3]{\frac 14\cos \bigg(\frac{6\pi}{17}\bigg) +\frac 34\sqrt[3]{\frac 14\cos \bigg(\frac{6\pi}{17}\bigg) +\frac 34\sqrt[3]{\frac 14\cos \bigg(\frac{6\pi}{17}\bigg) +\cdots}}}$$