# $\tan\frac{\pi}{9} +4\sin\frac{\pi}{9} = \sqrt 3$ [duplicate]

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?

• Does this help math.stackexchange.com/questions/3370381/… ?
– mark
Commented Jul 29, 2021 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
Commented Jul 29, 2021 at 21:17
• @islamm So there was a duplicate, strange that it didn’t come up either in this site’s search nor in Approach$0$. Commented Jul 29, 2021 at 21:21
• @KCd How is it incorrect? Commented Jul 29, 2021 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
Commented Jul 29, 2021 at 21:24

Here's a purely trigonometric solution, in case you're interested: $$\tan 20°+4\sin 20°=\frac {\sin 20°+2(2\sin 20° \cos 20°)}{\cos 20°}=\frac {(\sin 20°+\sin 40°)+\sin 40°}{\cos 20°}=\frac {2\frac 12 \cos 10°+\cos 50°}{\cos 20°}=\frac {2\cos 30°\cos 20°}{\cos 20°}=2 \frac {\sqrt 3}{2}=\sqrt 3$$
• Smooth! ${}{}{}$ Commented Jul 29, 2021 at 21:24