An interesting trigonometry problem -- featuring roots of unity.

We present a derivation of a pretty classic trigonometric identity that involves multiples of pi/11.

This is adapted from a solution in the Problems and Solutions section of the September 2006 issue of The College
Mathematics Journal.

[blackpenredpen]

Definitely the hardest and the most elegant trig expression I have seen

[aradhya]

But how do you discover these tools on your own when solving the problem? Using the tools is straightforward but how do you figure it out?

[felipelopes]

An explicit calculation consequence of the Kronecker-Weber theorem.
Quadratic extensions of Q are abelian, therefore are subfields of the n-th roots of unity for some n. This implies that some algebraic combination of the roots of unity should collapse into a quadratic rational :)

[Nithesh]

Great video! Sidenote: Deriving the second and third preliminaries can be done in a much quicker fashion if one uses the following identities (which follows from the even and odd properties of trig functions)

sin(x) = (e^ix - e^(-ix))/2i
cos(x) = (e^ix + e^(-ix))/2
itan(x) = (e^ix - e^(-ix))/(e^ix + e^(-ix)) (follows from the above two)

Then identity 2 is done by noticing that ζ^10 = ζ^-1 and identity 3 is done by noticing that (ζ^3 - 1)/(ζ^3 + 1) = (ζ^(3/2) - ζ^(3/2))/(ζ^(3/2) + ζ^-(3/2)).

[nasim]

And that's a good place to stop. :D That is always my favorite part of the video.
Thanks for the awesome problem-solving techniques, Michael!
Cheers!

[Adam-wmys]

You are the best at making me consider things I would've not thought of otherwise on such topics! Thank you 😁💗

[binaryblade]

if you factor exp(3pi/11) out of both your number and dem of the tan identity things simply quicker

[willyh.r.]

Aesthetic and technique in play. This is a Math ECSTASY !!! Thank you Michael Penn.

[nianli]

An easier way to find the solution:
Step 1 express the function in terms of sine and cosine functions only and then combine the two terms in P/Q form. Step 2 square the function to P^2/Q^2. Step3 use the elementary trigonometric formulas to express P^2 in a sum of cosine functions and a constant. Step4 after using tri. formulas again to sum cos 2x +cos 4x +cos 6x + cos 8x + cos 10x where x=Pi/11 you can obtain P^2=11Q^2。

[benardolivier]

Can this be generalized? For example, is there a, b and c such that tan(a*pi/n) + b*sin(c*pi/n) = sqrt(n) for any n?

[DrQuatsch]

I'd have multiplied top and bottom with e^(-3pi/11) for the itan(3pi/11) proof. That leads to [e^(3pi/11) - e^(-3pi/11)]/[e^(3pi/11) + e^(-3pi/11)] which is equal to 2isin(3pi/11)/2cos(3pi/11) which is itan(3pi/11).

[granhermon]

this should be into the overkill series

[malawigw]

I might do this with my class later this week, we just did Eulers formula today :)

[CTJ]

what is the character your using ? it sounds like ex-cee .. is that greek?

[Kishibe]

Is there any characterization of the linear combination of trigonometric functions and angles that such strategy resolves into closed form values?

[jeremygalvagni]

That's an interesting smartboard.

[alexpotts]

The "we can replace this 1 with a xi^33" step is among the biggest-brained ideas I have ever seen. Like, who would ever think to do that?

[neilgerace]

Can't get past 12:06, it keeps hanging ...?

[rhythmmandal]

we had AB and A+B why not do (A-B)^2 =

[beanhwak]

simpler o multiply n and d by e^-3teta