<s>[INST] Is it possible to express $\sin \frac{\pi}{9}$ in terms of radicals? So, yes, this is a math homework question. I've done some research on it and I know that the actual value for $\sin \frac{\pi}{9}$ cannot be expressed without using imaginary numbers.
http://intmstat.com/blog/2011/06/exact-values-sin-degrees.pdf
But, this isn't what the question is asking. It is simply asking if it is possible to do so and for me to prove it. I know that $\frac{3\pi}{9}$ can be simplified to $\frac{\pi}{3}$ and that exact values for the sine and cosine of it can be expressed cleanly and other multiples that can be reduced down to $\frac{\pi}{6}$, $\frac{\pi}{4}$, $\frac{\pi}{3}$ etc. 
But how can I prove that $\frac{\pi}{9}$ itself can be expressed as an exact value?
I'm in grade 12 advanced functions and am taking calculus next semester, but I'm totally open to learning new things so if you post very advanced concepts I'll do my best to understand them.
Any ideas where I could start?
Unit circle:
 [/INST] $$\sum_{k=1}^{\infty} \frac{1}{6}(-\frac{1}{2})^{k} = \frac{1}{9}\\$$
The basic unit circle goes all the way up to integral multiples of $\frac{\pi}{6}$; thus, it is reasonable to assume that $\sin{\frac{\pi}{9}}$ can reasonably be written as an infinite nested root through half-angle formulas. The resulting numerators of the above stated geometric sequence follow the pattern such that, assuming $a_{0}$ is 1, and ${k}$ follows suit with ${n}$:
\begin{align*}
a_{n+1} = 2a_{n}-1 \ \ \textrm{if} \ \ {n}\mod{2} = 0\\
a_{n+1} = 2a_{n}+1 \ \ \textrm{if} \ \ {n}\mod{2} = 1
\end{align*} 
Because of this, every third term ($a_{2},a_{5},...$) can be simplified, and thus have a slightly different final nested root.
Starting the half-angle chains using the sums produced from the equation above, it looks something like this:
\begin{gather*}
\sin{\frac{\pi}{6}}=\frac{1}{2}\\
\sin({\frac{\pi}{6}-\frac{\pi}{12}})=\sin{\frac{\pi}{12}}=\sqrt{\frac{1-\cos{\frac{\pi}{6}}}{2}}=\frac{\sqrt{2-\sqrt{3}}}{2}\\
\sin({\frac{\pi}{6}-\frac{\pi}{12}+\frac{\pi}{24}})=\sin{\frac{3\pi}{24}}=\sqrt{\frac{1-\sqrt{\frac{1+\cos{\frac{3\pi}{6}}}{2}}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}\\
\sin({\frac{\pi}{6}-\frac{\pi}{12}+\frac{\pi}{24}-\frac{\pi}{48}})=\sin{\frac{5\pi}{48}}=\sqrt{\frac{1-\sqrt{\frac{1+\sqrt{\frac{1+\cos{\frac{5\pi}{6}}}{2}}}{2}}}{2}}=\frac{\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{3}}}}}{2}\\
\end{gather*}
$\frac{3\pi}{24}$ is left as is to visibly see the divisibility of the fraction, and the correlation with both the terms before and after it regarding the nested roots.  Every 3 terms after this can also be simplified; thus, after the following terms of the sequence, things will start to repeat...
$$\sin({\frac{\pi}{6}-\frac{\pi}{12}+\frac{\pi}{24}-\frac{\pi}{48}+\frac{\pi}{96}})=\sin{\frac{11\pi}{96}}=\sqrt{\frac{1-\sqrt{\frac{1+\sqrt{\frac{1+\sqrt{\frac{1+\cos{\frac{11\pi}{6}}}{2}}}{2}}}{2}}}{2}}=\frac{\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{3}}}}}}{2}$$
After this term follows another value, $\frac{21\pi}{192}$, which can be simplified. Just like its predecessor, its final nested root is $\sqrt{2}$, which is then further nested similarly to the last occurrence. Since this is a recurring pattern, the value of $\sin{\frac{\pi}{9}}$ is thus defined to be
$$\frac{\sqrt{2-\Big[{\sqrt{2+{\sqrt{2-{\sqrt{2+...}}}}}}\Big]}}{2}$$
where the bracketed expression is self-recursive (that is, the expression fills in the ellipsis and leaves another ellipsis to be filled again). This value is approximately equal to $0.34202014333$.
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