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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

Henry Swanson
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user113528
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  • Under what conditions do you call a value exact? – Michael Hoppe Dec 05 '13 at 00:45
  • 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. – Stefan Smith Dec 05 '13 at 00:47
  • Yes, sorry for the confusion, the latter is what I am asking. How could I reword my question to cause less confusion? – user113528 Dec 05 '13 at 00:53
  • Hey, cool ... Your user number is an anagram of the first few terms of the Fibonacci sequence! :) – Blue Dec 05 '13 at 01:04
  • @Blue I'm not sure what you mean. I could do sin inverse to the left side but I don't think that's what you mean. – user113528 Dec 05 '13 at 01:08
  • While we can always use double angle formulae to readily evaluate $\text{sin}(\frac{\pi}{2^k})$ and $\text{sin}(\frac{\pi}{3\cdot 2^k}) $. But there is no way to solve this without adding some 'complexity' to it - no pun intended. – Christopher K Dec 05 '13 at 01:14
  • [ahem-hem] Writing $\theta$ for $\pi/9$, we have $$\sin 3\theta = \sin\frac{\pi}{3} = \sqrt{3}/2$$ What can you do to the left-hand side of this relation? (Hint: Express $\sin 3\theta$ in terms of $\sin\theta$.) – Blue Dec 05 '13 at 01:15
  • @Blue do you mean sin30=opposite/hypotenuse? – user113528 Dec 05 '13 at 01:24
  • @user113528: Nope. I mean like $\sin 2\theta = 2\sin\theta\cos\theta$ ... except for $\sin 3\theta$. – Blue Dec 05 '13 at 01:27
  • @Blue sin(pi/9)=2cos(pi/18)sin(pi/18) ? – user113528 Dec 05 '13 at 01:38
  • @user113528: That's true, but not what I'm suggesting. Expand $$\sin 3\theta = \sin(2\theta+\theta) = \sin 2\theta \cos\theta + ...$$ and keep going until you have nothing but $\sin\theta$ terms. – Blue Dec 05 '13 at 01:41
  • @Blue so sin(2θ+θ) expands to 2sinθcosθ + sinθ – user113528 Dec 05 '13 at 01:50
  • @user113528: $$\sin(2\theta+\theta) = \sin 2\theta \cos\theta + \cos 2\theta \sin\theta = (...)\cos\theta + (...)\sin\theta = ...$$ – Blue Dec 05 '13 at 02:05
  • edit:nevermind still working on it – user113528 Dec 05 '13 at 02:07
  • @Blue (2sinxcosx)cosx + (cos^2x-sin^2x)sinx okay I see it expands to sin3x=3sinxcos^2x−sin^3x but I'm not one hundred percent sure why it is we're using sin3θ in the first place? – user113528 Dec 05 '13 at 02:22
  • @user113528: The idea is to relate $\sin(\pi/9)$ to something (anything!) we already know. We know the trig values of $\pi/3$, and since it turns out that $\pi/3$ is (conveniently) a multiple of $\pi/9$, we go from there. – Blue Dec 05 '13 at 02:29
  • @Blue okay so now I get that too. My last question would be how did Tim Ratigan derive 64x^6−96x^4+36x^2−3 ? – user113528 Dec 05 '13 at 02:33
  • @user113528: $$\frac{\sqrt{3}}{2} = 3 x - 4 x^3 \quad\implies\quad \sqrt{3} = 6 s - 8 s^3$$ Now, square both sides. This has the cosmetic effect of eliminating the $\sqrt{3}$, so that all the coefficients are integers, but that's not really necessary here. You can apply the Cubic Formula to the un-squared polynomial equation just fine. (It'll be messy either way.) – Blue Dec 05 '13 at 02:43
  • @Blue where did the 3x-4x^3 come from? I thought we were at 3sinxcos^2x−sin^3x – user113528 Dec 05 '13 at 02:48
  • Re-write $\cos^2 x$. :) – Blue Dec 05 '13 at 02:59
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    @Blue alright thanks a lot for all your help! I definitely have a better understanding now. – user113528 Dec 05 '13 at 03:14
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    http://www.maths.manchester.ac.uk/~khudian/Etudes/Galetudes/galchildr1.pdf – lab bhattacharjee Dec 05 '13 at 03:52
  • @Blue I thought I had it but I don't. I still don't really understand where the 3x-4x^3 came from. – user113528 Dec 05 '13 at 04:59
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    @user113528: $$\sin 3\theta = 3 \sin\theta\cos^2\theta - \sin^3\theta = 3\sin\theta(1-\sin^2\theta) - \sin^3\theta = 3\sin\theta - 4\sin^3\theta$$ – Blue Dec 05 '13 at 05:12

3 Answers3

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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.

Tim Ratigan
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  • 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? – user113528 Dec 05 '13 at 01:29
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    @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). – Blue Dec 05 '13 at 01:37
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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 $0.34202014333$.

Steelscale
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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.

GEdgar
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    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. – Tim Ratigan Dec 05 '13 at 05:07
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    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. – GEdgar Dec 05 '13 at 14:15
  • $$\sin \left(\frac{\pi }{9}\right)=\frac{i \left(\sqrt[3]{1-i \sqrt{3}}-\sqrt[3]{1+i \sqrt{3}}\right)}{2 \sqrt[3]{2}}$$ – Claude Leibovici Mar 07 '22 at 09:50