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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:Unit circle

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  • $\begingroup$ Under what conditions do you call a value exact? $\endgroup$ – Michael Hoppe Dec 5 '13 at 0:45
  • $\begingroup$ I'm not sure what your question is. $\pi/9$ is an exact value. It is $\pi$ times $1/9$. Like $\pi$, $\pi/9$ is a transcendental number : I suggest Googling "transcendental number" or finding a good book discussing the topic. If you are asking about whether $\sin(\pi/9)$ can be expressed in a different form, using square roots and fractions and such, then you need to reword your question. $\endgroup$ – Stefan Smith Dec 5 '13 at 0:47
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    $\begingroup$ @Blue alright thanks a lot for all your help! I definitely have a better understanding now. $\endgroup$ – user113528 Dec 5 '13 at 3:14
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    $\begingroup$ @user113528: $$\sin 3\theta = 3 \sin\theta\cos^2\theta - \sin^3\theta = 3\sin\theta(1-\sin^2\theta) - \sin^3\theta = 3\sin\theta - 4\sin^3\theta$$ $\endgroup$ – Blue Dec 5 '13 at 5:12
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By Euler's formula, $e^{xi}=\cos x+i\sin x$. Therefore $e^{3xi}=(\cos x+i\sin x)^3=\cos 3x+i\sin 3x$. Equating imaginary parts, $\sin 3x=3\sin x\cos^2x-\sin^3x$. Therefore:

$$\begin{align}\sin\frac\pi3&=3\sin\frac\pi9\cos^2\frac\pi9-\sin^3\frac\pi9\\ \frac{\sqrt{3}}2&=3\sin\frac\pi9-4\sin^3\frac\pi9\end{align}$$

This tells us $\sin\frac\pi9$ is a root of $64x^6-96x^4+36x^2-3$. If you're determined, you can try using the solution to the general cubic equation to find the value.

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  • $\begingroup$ Ok this is the first time I've ever seen Euler's formula and after doing a little reading I still don't really understand what you did it all. I'll continue doing reading but but can you explain to me why you changed e^{xi} to e^{3xi}? How did you equate the imaginary parts? And how does the final line you posted show us that (pi/9) is a root of that equation? $\endgroup$ – user113528 Dec 5 '13 at 1:29
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    $\begingroup$ @user113528: Euler's formula (more precisely, DeMoivre's Formula) is (very!) convenient (trust me ... you'll love it when you get to it), but it's not necessary. You can expand $\sin 3\theta$ by more elementary methods to get the second of Tim's displayed equations (which I would leave as $\sqrt{3}/2 = 3 s - 4 s^3$ before attempting to apply the Cubic Formula). $\endgroup$ – Blue Dec 5 '13 at 1:37
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In Maple, convert(RootOf(64*x^6-96*x^4+36*x^2-3,x),radical); has result:

$$ \frac{1}{4}\,\sqrt {-\sqrt [3]{4+4\,i\sqrt {3}}-4\,{\frac {1}{\sqrt [3]{4+4\,i \sqrt {3}}}}+8+8\,i\sqrt {3} \left( \frac{1}{8}\sqrt [3]{4+4\,i\sqrt {3}}-\frac{1}{2}\,{\frac {1}{\sqrt [3]{4+4\,i\sqrt {3}}}} \right) } $$

Of course in the "casus irreducibilis", even though the root is real, there are complex numbers involved in the calculation.

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    $\begingroup$ I almost posted this (using Mathematica), but I think you'll find that if you simplify this algebraically, it reduces to $\sin \frac\pi9$ somewhat tautologically only by assuming De Moivre; otherwise you're just left with complex radicals. $\endgroup$ – Tim Ratigan Dec 5 '13 at 5:07
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    $\begingroup$ In the casus irreducibilis, you cannot use only real radicals; you can use either complex radicals, or trig functions. Since the question is about radicals, I posted this. $\endgroup$ – GEdgar Dec 5 '13 at 14:15
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$$\sum_{k=1}^{\infty} \frac{1}{6}(-\frac{1}{2})^{k} = \frac{1}{9}\\$$ 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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