# Ask Uncle Colin: A hellish trigonometric identity

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

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:

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

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

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

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

It gets nicer, honestly!

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

This is the same as

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

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:

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

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

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

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