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

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

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

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beautiful! it seems there is no question i know which you cannot solve !
Andre thats make me cry :\ @adxpoi lol
Haha, nah I'm sure there's stuff I can't solve :P
extremely interesting question hmmm and answer :O
Hey I'm pretty sure I've asked this one before
I think @adxpoi 's method generalizes to any arbitrary degree polynomial : Find a polynomial of \(n\)th degree whose roots are \[\cot^2\frac{\pi}{2n+1},~\cot^2\frac{2\pi}{2n+1},~\cot^2\frac{3\pi}{2n+1},\cdots, \cot^2\frac{n\pi}{2n+1}\]
Yes @ParthKohli I remember, we had solved it using Vieta's formulas or something adhoc... :)
No, complex numbers.
In fact just like how adx has solved it.
Ahh okay, can't tryst my memory lol knw how to solve the general case for arbitrary degree ?
My book (KD Joshi) has this, yeah.
Ohk.. I'm hoping we can simply mimic the adx's method, but I haven't tried it yet... so not completely sure yet..
Where did you find this question?
IMO shortlisted practice problems
@praxer has been asking these recently
Hmm, this one is pretty standard for an IMO problem if you've read about it.
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
Exactly.
Is there any other way to see that \(e^{i2k\pi/7}\) satisfies that polynomial equation ? (w/o using geometric series)
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).
No, not really.
Pages 246-247, yes.
slightly different problem, but the methods are similar yeah https://i.gyazo.com/27c20454519da07d956f53137adc3ee6.png

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