Question

find a third degree polynomial with integer coefficients whose roots are \[\cot^2\frac{\pi}{7},~\cot^2\frac{2\pi}{7},~\cot^2\frac{3\pi}{7}\]

Answer 1

the sum and product of roots do give rational numbers

http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D1%7D%5E3+%28cot%28k*pi%2F7%29%29%5E2

http://www.wolframalpha.com/input/?i=%5Cprod%5Climits_%7Bk%3D1%7D%5E3+%28cot%28k*pi%2F7%29%29%5E2

Answer 2

Consider the polynomial equation:
\(x^6 +x^5 + x^4 + x^3 + x^2 + x + 1 = 0\)

It's roots are \(e^{i2k\pi/7}\) where \(k = 1\) to \(6\)

Divide the equation by \(x^3\) , then we have,
\(x^3 + x^2 + x + 1 + 1/x + 1/x^2 + 1/x^3 = 0 \)

\( (x+1/x)^3 -3(x+1/x) + (x+1/x)^2 - 2 + (x+1/x) + 1 = 0 \)

When x = \(e^{i2k\pi/7}\) , \((x+1/x) = 2\cos(2k\pi/7)\)

So, \(y^3 + y^2 -2y -1 = 0\) has the roots \(2\cos(2\pi/7), 2\cos(4\pi/7) , 2\cos(6\pi/7)\)

Now, note that \( \cos(2k\pi/7) = \cos^2(k\pi/7) - \sin^2(k\pi/7)\)

\(= \frac{1}{\sec^2(k\pi/7)} - \frac{\tan^2(k\pi/7)}{\sec^2(k\pi/7)}\)

\(= \frac{1-\tan^2(k\pi/7)}{1 + \tan^2(k\pi/7)} \)

\(= \frac{\cot^2(k\pi/7) - 1}{\cot^2(k \pi/7) + 1}\)

So, if \(\cot^2(k\pi/7) = t\) , \(2\cos(2k\pi/7) = 2\frac{(t-1)}{(t+1)}\)

Therefore \(\cot^2(k\pi/7)\) satisfies the following equation

\( (2\frac{(t-1)}{(t+1)})^3 + (2\frac{(t-1)}{(t+1)})^2 -2(2\frac{(t-1)}{(t+1)}) - 1 = 0\)

Multiplying both sides by \((t+1)^3\) ,
\( (2t-2)^3 + (t+1)(2t-2)^2 -2(t+1)^2(2t-2) - (t+1)^3 = 0\)

Therefore,
\( (2t-2)^3 + (t+1)(2t-2)^2 -2(t+1)^2(2t-2) - (t+1)^3 \)
is clearly a polynomial with integer coefficients and has as its roots \(\cot^2(k\pi/7)\) for \(k = 1,2,3\)

Answer 3

Also, simplifying the final expression, the final polynomial turns out to be
\(7t^3 - 35t^2 + 21t-1\)

Answer 4

beautiful! it seems there is no question i know which you cannot solve !

Answer 5

Andre thats make me cry :\ @adxpoi lol

Answer 6

Haha, nah I'm sure there's stuff I can't solve :P

Answer 7

extremely interesting question hmmm and answer :O

Answer 8

Hey I'm pretty sure I've asked this one before

Answer 9

I think @adxpoi 's method generalizes to any arbitrary degree polynomial :

Find a polynomial of \(n\)th degree whose roots are
\[\cot^2\frac{\pi}{2n+1},~\cot^2\frac{2\pi}{2n+1},~\cot^2\frac{3\pi}{2n+1},\cdots, \cot^2\frac{n\pi}{2n+1}\]

Answer 10

Yes @ParthKohli I remember, we had solved it using Vieta's formulas or something adhoc... :)

Answer 11

No, complex numbers.

Answer 12

In fact just like how adx has solved it.

Answer 13

Ahh okay, can't tryst my memory lol
knw how to solve the general case for arbitrary degree ?

Answer 14

My book (KD Joshi) has this, yeah.

Answer 15

Ohk.. I'm hoping we can simply mimic the adx's method, but I haven't tried it yet... so not completely sure yet..

Answer 16

Where did you find this question?

Answer 17

IMO shortlisted practice problems

Answer 18

@praxer has been asking these recently

Answer 19

Hmm, this one is pretty standard for an IMO problem if you've read about it.

Answer 20

I read and forget most of what I read, so I must have encountered this problem before.. but I'm pretty sure I never had this observation before :

\[x^6+x^5+\cdots+x+1 = \dfrac{x^7-1}{x-1}\]
so \(e^{i2k\pi/7}\) is a root

Answer 21

Exactly.

Answer 22

Is there any other way to see that \(e^{i2k\pi/7}\) satisfies that polynomial equation ? (w/o using geometric series)

Answer 23

Here's a direct-download link to the book I'm talking about:

https://doc-0o-44-docs.googleusercontent.com/docs/securesc/ha0ro937gcuc7l7deffksulhg5h7mbp1/l4j8b8u0i1870es3h7u201vqlntqv62q/1439287200000/06023952678462293309/*/0BwEaIp7y7BJTOVZVcVZiVUw3aXM?e=download

The author explains the general case really nicely in the trigonometric identities or equations chapter (I don't remember which one it was).

Answer 24

No, not really.

Answer 25

Pages 246-247, yes.

Answer 26

slightly different problem, but the methods are similar yeah
https://i.gyazo.com/27c20454519da07d956f53137adc3ee6.png