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Prove that $$ \tan\frac{\pi}{9} +4\sin\frac{\pi}{9} = \sqrt 3 $$

There seem to be a lot of similar identities that are provable, for example, by using roots of unity. However, here I cannot get things to work out nicely.

If $u=e^{\frac{2\pi i}{9}} $, then $$i\left(\tan\frac{\pi}{9} +4\sin\frac{\pi}{9}\right) =\frac{u-1}{2(u+1)} +2(u^4-u^5)=\frac{-4u^6 +4u^4+u-1}{2(u+1)} $$ and so $$\left(\tan\frac{\pi}{9} +4\sin\frac{\pi}{9}\right)^2 = 3 \\ \iff (-4u^6+4u^4+u-1)^2+12(u+1)^2 =0 \\ \iff 16u^8-8u^7+8u^6+8u^5-8u^4+16u^3+13u^2-10u+13 =0$$ Unfortunately, the LHS is not of the form $k(u^8+u^7 +\dots+1)$ making the equality unobvious. How to proceed?

Vishu
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    Does this help https://math.stackexchange.com/questions/3370381/prove-that-tan-frac-pi9-4-sin-frac-pi9-sqrt3 ? – mark Jul 29 '21 at 21:16
  • You should check your algebraic calculations from the start, since the first line after you define $u$ does not appear to be correct. – KCd Jul 29 '21 at 21:17
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    @islamm So there was a duplicate, strange that it didn’t come up either in this site’s search nor in Approach$0$. – Vishu Jul 29 '21 at 21:21
  • @KCd How is it incorrect? – Vishu Jul 29 '21 at 21:21
  • @Tavish when I substitute $e^{2\pi i/9}$ for $u$ in the rational function $(-4u^6 +4u^4+u-1)/(2(u+1))$, I do not get a value that is nearly $\sqrt{3}i$. Do you? – KCd Jul 29 '21 at 21:24
  • @KCd I’ve checked my algebra and it’s correct. What calculator are you using? – Vishu Jul 29 '21 at 21:28
  • If your algebra is correct then $(-4u^6 + 4u^4 + u-1)/(2(u-1)) = \sqrt{3}i$ when $u = e^{2\pi i/9}$. At that value of $u$, I find the ratio is $1.550065\ldots i$, which is purely imaginary but is not $\sqrt{3}i$. What do you find numerically for the value at $e^{2\pi i/9}$? – KCd Jul 29 '21 at 21:36
  • @KCd Hmm, I get $1.55 i $ too, but I can’t find a mistake in my calculations. Can you find one? – Vishu Jul 29 '21 at 21:50
  • The only thing I did was plug in numbers for $u$, and since I did not get $\sqrt{3}i$ (and I redid that calculation again to be sure of it) I knew you had an error at that point or earlier. There is only one previous step, and you are the one who has your calculations written down, so you should look at what you wrote to find out why you did not correctly express $\tan(\pi/9) + 4\sin(\pi/9)$ in terms of $u$. – KCd Jul 29 '21 at 21:55
  • @KCd I know either the error is from the computer or my step, and of course I have looked at my step and can’t seem to find an error, which is why I asked. – Vishu Jul 29 '21 at 22:00
  • Nobody but you can see your work, so there's no way I or anyone else can figure out where you made a mistake in the conversion of the numerical sine and tangent expression into an algebraic expression. Your very first formula in terms of $u$ is an incorrect expression for the sine and tangent expression when $u = e^{2\pi i/9}$, so you're just going to have to work out your error by yourself or just forget it. – KCd Jul 30 '21 at 03:45
  • @KCd It isn’t true that only I can detect an error, even if the work is mine. I have to say again, I didn’t find any error, so someone else is going to have to tell me where there is one. – Vishu Jul 30 '21 at 11:32
  • This will be my final comment. Nobody else can find your error since you give no work where the error was made. Near the start you say $A = B$ (where $A = i(\tan(\pi/9) + 4\sin(\pi/9))$ and $B = (u-1)/(2(u+1)) + 2(u^4 - u^5)$ for $u = e^{2\pi i/9}$) but there are no steps showing how you got from $A$ to $B$. The mistake is obviously there since it is easy to check $A \not= B$ numerically, but since there is nothing shown about your process of going from $A$ to $B$, you are being unreasonable to expect other people can find the mistake. – KCd Jul 30 '21 at 13:52
  • @KCd You have finally answered my question. So you say the error was made at that line. All I did there was use the fact that $\tan\frac{\pi}{9} = \frac{u-1}{2i(u+1)} $ and $\sin\frac{\pi}{9} = \frac{u^4-u^5}{2i} $ since $u^4 = -e^{-i\pi/9} $ and $u^5 = - e^{i \pi/9} $. Now, have I stated anything wrong here? – Vishu Jul 30 '21 at 14:31
  • I had already told you before that the error was there when I wrote "your very first formula in terms of $u$ is an incorrect expression for the sine and tangent expression when $u = e^{2\pi i/9}$." You are on your own now. I am not continuing this discussion. – KCd Jul 30 '21 at 14:35
  • @KCd You’re just repeating the same thing, which is not helpful anyway. You haven’t answered my question in my last comment. – Vishu Jul 30 '21 at 14:41
  • @Vishu, this answer of mine may be of your interest. – Sahaj Dec 24 '23 at 12:30

1 Answers1

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Here's a purely trigonometric solution, in case you're interested: $$\tan 20°+4\sin 20°=\frac {\sin 20°+2(2\sin 20° \cos 20°)}{\cos 20°}=\frac {(\sin 20°+\sin 40°)+\sin 40°}{\cos 20°}=\frac {2\frac 12 \cos 10°+\cos 50°}{\cos 20°}=\frac {2\cos 30°\cos 20°}{\cos 20°}=2 \frac {\sqrt 3}{2}=\sqrt 3$$

Ritam_Dasgupta
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