Ask Uncle Colin: A hellish trigonometric identity

Dear Uncle Colin,

@CmonMattTHINK unearthed the challenge to prove that:

$\tan \left(\frac{3}{11}\pi \right)+4\sin \left(\frac{2}{11}\pi \right)=\sqrt{11}$. Wolfram Alpha says it’s true, but I can barely get started on the proof and I’m worried no-one will like me.

Grr, Really Obnoxious Trigonometry Has Evidently Nice Denouement – Is Everything Common Knowledge?

Hi, GROTHENDIECK – don’t worry, there is practically no correlation between one’s ability to do trig proofs and one’s ability to make friends.

This particular problem is a Big Blue Meany. It took me a couple of days to get anywhere near a solution, too, and even afterwards I’m not 100% happy with it. (There’s a paper on it here, though, with some alternatives if you don’t like mine.)

My working approach is to turn everything into complex form, letting $z=e^{\frac{1}{11}\pi }$ for convenience:

$\tan \left(\frac{3}{11}\pi \right)\equiv \frac{z^3-z^{-3}}{i(z^3+z^{-3})}$

$4\sin \left(\frac{2}{11}\pi \right)\equiv 2\frac{z^2-z^{-2}}{i}$

Doesn’t look much nicer, does it? OK, let’s add the fractions to get :

$\frac{z^3-z^{-3}}{i(z^3+z^{-3})}+2\frac{z^2-z^{-2}}{i}=\frac{z^3-z^{-3}+2(z^2-z^{-2})((z^3+z^{-3})}{i(z^3+z^{-3})}$

It gets nicer, honestly!

$\frac{z^3-z^{-3}+2(z^5+z-z^{-1}-z^{-5})}{i(z^3+z^{-3})}$

This is the same as

$\frac{2z^5+z^3+2z-2z^{-1}-z^{-3}-2z^{-5}}{i(z^3+z^{-3})}$

Square and square root:

$\pm \sqrt{\frac{4z^{10}+4z^8-7z^6+4z^4+4z^2-18+4z^{-2}+4z^{-4}-7z^{-6}+4z^{-8}+4z^{-10}}{-(z^6+2+z^{-6})}}$

No, wait, come back! Remember the first rule of roots of unity club? No? Well, you should! The eleventh roots of unity, that is to say $z^{-10}+z^{-8}+z^{-6}+z^{-4}+z^{-2}+1+z^2+z^4+z^6+z^8+z^{10}=0$. Turns out, we’ve got nearly all of those – although we’ll need to add and subtract $4z^6$, $4z^{-6}$ and $4$ to make it work properly:

$\pm \sqrt{\frac{[4z^{10}+4z^8+4z^6+4z^4+4z^2+4+4z^{-2}+4z^{-4}+4z^{-6}+4z^{-8}+4z^{-10}]-11z^6-22-11z^{-6}}{-(z^6+2+z^{-6})}}$

The big bracket is zero:

$\pm \sqrt{\frac{-11z^6-22-11z^{-6}}{-(z^6+2+z^{-6})}}$

… which is clearly $\pm \sqrt{11}$. However, since both of our original trig function are in the first quadrant, our answer must be the positive root.

Incidentally, GROTHENDIECK? Asking people to solve that sort of puzzle is a sure-fire way to lose friends.

-- Uncle Colin