cos k theta and cos(k + 1) theta are both rational

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#1 h Dec 8, 2007, 1:39 AM • 4 Y

Y by Davi-8191, Adventure10, Mango247, cubres

Let $\theta$ be an angle in the interval $(0,\pi/2)$ . Given that $\cos \theta$ is irrational, and that $\cos k \theta$ and $\cos[(k + 1)\theta ]$ are both rational for some positive integer $k$ , show that $\theta = \pi/6$ .

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#2 h Dec 25, 2007, 5:04 AM • 3 Y

Y by Adventure10, Mango247, cubres

Here's an idea:
We can prove easily by induction, that for each integer $k$ there exists a polynomial $R_{k}$ so that $R_{k} (cos x) = cos (kx)$ .
Suppose $x = cos \theta$ is irrational. Then $R_{k} (x) = - p$ and $R_{k + 1} (x) = - q$ are both rational. Thus $R_{k} (x) + p$ and $R_{k + 1} (x) + q$ are polynomials with rational coefficients, which have zeros at $x$ . Consider the monic polynomial with rational coefficients $Q$ of minimal degree such that $S(x) = 0$ . Using the well-known Lemma:[ if for an algebraic number $x$ the rational polynomial of minimal degree which has $x$ as a 0 is $Q(x)$ , then if $R(x) = 0$ for some rational polynomial $R(x)$ , we have $Q(x)$ is a factor of $R(x)$ .], we have $Q(x)$ is a factor of both $R_{k} (x) + p$ and $R_{k + 1} (x) + q$ , so $R(x)$ is a factor of $(R_{k} (x) + p, R_{k + 1} (x) + q)$ .
Since $x$ is irrational, the degree of $Q(x)$ is at least $2$ . Our goal is to prove that the degree of $(R_{k} (x) + p, R_{k + 1} (x) + q)$ , and hence $Q(x)$ is $2$ . To do that, it may be possible to apply Euclidean algorithms to those 2 polynomials. Hopefully, there will be a special relationship between the coefficients of the polynomials that we obtain, which will make this easier to do. I think we can prove, using this method that the degree of $(R_{k} (x) + p, R_{k + 1} (x) + q)$ is at most $2$ , and if it is $2$ then $p$ and $q$ must take specific values (depending on the value of $k (mod 6)$ ). If $k \equiv 1 (mod 3)$ , using this method I think we should get a contradictory relationships between $p,q$ , if the degree of the LCD polynomial is 2, that means $k \equiv 1 (mod 3)$ is not possible. Once, for a certain value of $k$ , we find out what $p$ and $q$ are (one should be $0.5$ or $- 0.5$ , the other should be $0$ ), the problem will be finished.
----
Here's how we can kill $k = 3$ using this method:
Since $P_{3} (x) = 4x^3 - 3x$ and $P_{4} (x) = 8x^4 - 8x^2 + 1$ , let $4x^3 - 3x + a = 0, 8x^4 - 8x^2 + b = 0$ for rational $a,b$ . Then if $Q(x)$ is minimal integer polynomial of $x$ , $Q(x)$ divides $4x^3 - 3x + a$ and $8x^4 - 8x^2 + b$ , $Q(x)$ has degree at least $2$ . We have:
$(4x^3 - 3x + a, 8x^4 - 8x^2 + b) = (4x^3 - 3x + a, 2x^2 + 2ax - b) = (2x^2 + 2ax - b, 4ax^2 - (2b - 3) x - a)$ .
Here we have used typical Euclidean algorithm: using fact that $8x^4 - 8x^2 + b = (4x^3 - 3x + a)2x - (2x^2 + 2ax - b)$ , etc. Since the GCD of those 2 polynomials, degree $2$ , is a polynomial of degree $2$ , either the $2$ -nd one is $0$ - giving us $b = 3/2, a = 0$ from where we get that $cos (3 \theta) = 0$ , $cos (4 \theta) = - 0.5$ and $\theta = \frac {\pi}{6}$ as required: OR the second quadratic is a rational multiple of the first. In this case, a routine calculation gives a contradic tion.

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#3 h Jan 12, 2008, 7:14 PM • 3 Y

Y by Adventure10, Mango247, cubres

N.T.TUAN, excuse me, where did you find USA TST 2007?

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#4 h Feb 11, 2008, 7:39 AM • 3 Y

Y by Adventure10, Mango247, cubres

cheese: they've been posted here: http://www.mathlinks.ro/viewtopic.php?t=177771

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#5 h Feb 11, 2008, 10:45 AM • 2 Y

Y by Adventure10, cubres

Hm.. in last year, I saw this problem and tried to solve this problem by same method as yours.... I checked it for $k=3,4,5$ but I failed to prove for all cases.
in my case, It isn't enough to prove it, but it needs more facts such as $|x|\le 1$ ...., please post details.. I want to see a solution of this problem.

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#6 h Feb 12, 2008, 6:38 AM • 3 Y

Y by Adventure10, Mango247, cubres

I don't know why I couldn't think about this......
Let $T_{n}$ be a integer coefficient polynomial such that $T_{n}(\cos \theta) = \cos n\theta$ .
It is easy to prove that $T_{n + 1}(x) + T_{n - 1}(x) = 2xT_{n}(x)$ ,
Therfore $deg( T_{n}) = n$ , $[x^{n}]T_{n} = 2^{n - 1}$ , $[x^{n - 2}]T_{n} = - 2^{n - 3}n$ .........

Let $T_{k}(\cos \theta) = p$ , $T_{k + 1}(\cos \theta) = q$ . It is easy to prove that
if $R(x) = x^{2} - 2pqx + p^{2} + q^{2} - 1$ , $R(\cos \theta) = R(\cos (2k + 1)\theta) = 0$ ....
It is obvious that $R$ is a minimal degree polynomial which has root $\cos \theta$ .

So $R(x)|T_{k}(x) - p, T_{k + 1}(x) - q$ ,
$\cos k\theta = \cos (2k^{2} + k)\theta = p$ ,
$\cos (k + 1)\theta = \cos (2k + 1)(k + 1)\theta = q$ ,

Let's Suppose that $\frac {k(k + 1)\theta}{\pi} \not\in \mathbb{Z}$ .
then $\frac {k^{2}\theta}{\pi}, \frac {(k + 1)^{2}\theta}{\pi} \in \mathbb{Z}$
$\Rightarrow \frac {(2k + 1)\theta}{\pi}\in\mathbb{Z} \rightarrow \cos (2k + 1)\theta = \pm 1$ , but $\cos \theta$ is irrational, contradiction!

Now $\frac {k(k + 1)\theta}{\pi} \in \mathbb{Z}$ .
$\Rightarrow T_{k}(q) = T_{k + 1}(p) = \pm 1$ .
$q = \frac {a}{b}$ ( $(a,b) = 1$ ). if $k\ge3$ , then we can easily show that $q = 0$ or $a = \pm 1$ and $b = \pm 2^{i}$ .
Therfore $p, q = \pm \frac {1}{2^{i}}$ or $0$ .

If one can prove that $T_{n}(\frac {1}{2^{i}})\not = \pm1$ if $i\ge 2$ , this is the end..
I think it isn't very hard.... but I don't have times to check it..

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#7 h Feb 22, 2009, 8:58 AM • 3 Y

Y by Adventure10, Mango247, cubres

Little Gauss wrote:

I don't know why I couldn't think about this......
Let $T_{n}$ be a integer coefficient polynomial such that $T_{n}(\cos \theta) = \cos n\theta$ .
It is easy to prove that $T_{n + 1}(x) + T_{n - 1}(x) = 2xT_{n}(x)$ ,
Therfore $deg( T_{n}) = n$ , $[x^{n}]T_{n} = 2^{n - 1}$ , $[x^{n - 2}]T_{n} = - 2^{n - 3}n$ .........

Let $T_{k}(\cos \theta) = p$ , $T_{k + 1}(\cos \theta) = q$ . It is easy to prove that
if $R(x) = x^{2} - 2pqx + p^{2} + q^{2} - 1$ , $R(\cos \theta) = R(\cos (2k + 1)\theta) = 0$ ....
It is obvious that $R$ is a minimal degree polynomial which has root $\cos \theta$ .

So $R(x)|T_{k}(x) - p, T_{k + 1}(x) - q$ ,
$\cos k\theta = \cos (2k^{2} + k)\theta = p$ ,
$\cos (k + 1)\theta = \cos (2k + 1)(k + 1)\theta = q$ ,

Let's Suppose that $\frac {k(k + 1)\theta}{\pi} \not\in \mathbb{Z}$ .
then $\frac {k^{2}\theta}{\pi}, \frac {(k + 1)^{2}\theta}{\pi} \in \mathbb{Z}$
$\Rightarrow \frac {(2k + 1)\theta}{\pi}\in\mathbb{Z} \rightarrow \cos (2k + 1)\theta = \pm 1$ , but $\cos \theta$ is irrational, contradiction!

Now $\frac {k(k + 1)\theta}{\pi} \in \mathbb{Z}$ .
$\Rightarrow T_{k}(q) = T_{k + 1}(p) = \pm 1$ .
$q = \frac {a}{b}$ ( $(a,b) = 1$ ). if $k\ge3$ , then we can easily show that $q = 0$ or $a = \pm 1$ and $b = \pm 2^{i}$ .
Therfore $p, q = \pm \frac {1}{2^{i}}$ or $0$ .

If one can prove that $T_{n}(\frac {1}{2^{i}})\not = \pm1$ if $i\ge 2$ , this is the end..
I think it isn't very hard.... but I don't have times to check it..

A wondeful approach

We needn't to prove $T_{n}(\frac {1}{2^{i}})\not = \pm1$ if $i\ge 2$
We just take your conclusion $\frac {k(k + 1)\theta}{\pi} \in \mathbb{Z}$
We get: $\theta$ is a rational multiple of $\pi$
From $R$ is a minimal degree polynomial which has root $\cos \theta$ yields $\cos \theta$ is a number whose algebraic degree is 2
From an old assertion of Hsiao (easy to prove)
$\theta = \frac {p}{q}\pi$ that $q = 4,5,6$
Easy to check $q\neq 4,5$ yields $q = 6,p = 1$ that finished our proof

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#8 h Aug 10, 2010, 11:22 PM • 8 Y

Y by v_Enhance, djmathman, Aryan-23, rayfish, Adventure10, Mango247, cubres, and 1 other user

The problem statement is still true if $k+1$ is replaced by an integer $l$ relatively prime to $k$ .

Lemma 1
The only rational $p\in [0,1]$ such that $\cos p\pi$ is rational are $p=0, \frac{1}{3},\frac{1}{2},\frac{2}{3},1$ .
Proof
Let $\omega=\cos (p\pi)+i\sin(p\pi)$ . Then $\omega$ is a root of unity and hence an algebraic integer; similarly $\overline{\omega}$ is as well. Hence $2\cos(p\pi) =\omega + \overline{\omega}$ is an algebraic integer; since it is rational it is a rational (i.e. regular) integer. Hence we must have $2\cos(p\pi)=0,\pm 1,\pm 2$ , giving $p=0, \frac{1}{3},\frac{1}{2},\frac{2}{3},1$ .

Lemma 2
If $\cos k\theta$ and $\cos l\theta$ are rational for relatively prime positive integers $k,l$ , then either $\cos \theta$ is rational or $\theta$ is a rational multiple of $\pi$ .
Proof
Let $\cos k\theta=p$ and $\cos l\theta =q$ . Noting that $e^{i\phi}+\frac{1}{e^{i\phi}}=2\cos \phi$ , we have that $e^{i\theta}$ is a root of
$\begin{align} x^k+\frac{1}{x^k}=2p&\implies x^{2k}-2px^k+1=0\\ x^l+\frac{1}{x^l}=2q&\implies x^{2l}-2qx^l+1=0 \end{align}$
Suppose $\theta$ is not a rational multiple of $\pi$ . Then the numbers $e^{\pm i\theta+\frac{2\pi i j}{k}}$ for $0\leq j<k$ are all distinct. They are the $2k$ roots of (1). Similarly, $e^{\pm i\theta+\frac{2\pi i j}{l}}$ are the $2l$ roots of (2). Since $\theta$ is not a rational multiple of $\pi$ and $k,l$ are relatively prime, we have that the only solution to $e^{\pm i\theta+\frac{2\pi i j_1}{k}}= e^{\pm i\theta+\frac{2\pi i j_2}{l}}$ for $0\leq j_1<k$ and $0\leq j_2<l$ for some choices of signs is when $j_1=j_2=0$ . Thus the only roots the two polynomials share in common are $\omega=e^{i\theta}$ and $\bar{\omega}=-e^{i\theta}$ , and the polynomial
$g(x):=(x-e^{i\theta})(x-e^{-i\theta})$
is the greatest common divisor of polynomials $x^{2k}-2px^k+1$ and $x^{2l}-2qx^l+1$ , which have rational coefficients. Hence $g(x)$ has rational coefficients, and $e^{i\theta}+e^{-i\theta}=2\cos \theta$ is rational.

By Lemma 2, $\theta$ is a rational multiple of $\pi$ . Since by Lemma 1 the only rational multiples of $\pi$ that give a rational values for cosine are multiples of $\frac{\pi}{6}$ , $k\theta$ and $l\theta$ are both multiples of $\frac{\pi}{6}$ . By Bezout's Theorem $\theta=m(k\theta)+n(l\theta)$ for some integers $m,n$ , so $\theta$ is a multiple of $\frac{\pi}{6}$ . Since $\cos\theta$ is irrational, $\theta=\frac{\pi}{6}$ .

Note: This solution is based on the idea of the solution to the problem here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=589&t=290391

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#9 h Aug 21, 2017, 2:58 AM • 3 Y

Y by pieater314159, Adventure10, cubres

Let $x = \cos \theta + i \sin \theta$ , and let $z_m = x^m + \tfrac{1}{x^m} = 2 \cos m\theta$ for all integer $m$ . We are told that $z_k$ and $z_{k + 1}$ are both rational but $z_1$ is not.

Let $T_n(z)$ be the monic polynomial with integer coefficients satisfying $T_n(2 \cos x) = 2 \cos nx$ for all $x$ .

Let $q(z) = \gcd(T_n(z) - z_n, T_{n + 1}(z) - z_{n + 1})$ . Then $q$ has rational coefficients and $z_1$ is a root of $q$ . Thus, if $\deg q = 1$ , then $z_1$ is rational, contradiction.

Thus, $\deg q \ge 2$ . If $q$ has root $z_1$ with multiplicity $\deg q$ , then it's easy to see $z_1$ is rational, contradiction. Thus, $q$ has a root $w \neq z_1$ .

Let $y$ be a number with $y + \tfrac{1}{y} = w$ . Then since $x + \tfrac{1}{x} = z_1 \neq w = y + \tfrac{1}{y}$ , $x \neq y$ and $xy \neq 1$ .

Now, $x^k + \tfrac{1}{x^k} = y^k + \tfrac{1}{y^k} = z_k$ and $x^{k + 1} + \tfrac{1}{x^{k + 1}} = y^{k + 1} + \tfrac{1}{y^{k + 1}} = z_{k + 1}$ . Thus, we obtain

$x^k = y^k$ or $(xy)^k = 1$ , and
$x^{k + 1} = y^{k + 1}$ or $(xy)^{k + 1} = 1$ .

By setting $y \to \tfrac{1}{y}$ if necessary, assume $x^k = y^k$ . Then $x^{k + 1} = y^{k + 1}$ implies $x = y$ , bad. Thus $(xy)^{k + 1} = 1$ .

Since $(x/y)^k = (xy)^{k + 1} = 1$ , both $x/y$ and $xy$ are roots of unity, so $x$ and $y$ are too. It follows that angles $\alpha_1 = k \theta$ and $\alpha_2 = (k + 1) \theta$ are a rational number of degrees and have rational cosines, so
$\cos k\theta, \cos (k + 1)\theta \in \{-1, -\tfrac{1}{2}, 0, \tfrac{1}{2}, 1\}.$ It follows that $k \theta$ and $(k + 1) \theta$ are both multiples of $30^{\circ}$ , so $\theta$ is too.

Since $\theta \in (0^{\circ}, 90^{\circ})$ , we obtain $\theta = 30^{\circ}$ or $\theta = 60^{\circ}$ . Since $\cos 60^{\circ}$ is rational, it follows that $\theta = 30^{\circ}$ .

The solution also works for the following problem:

Quote:

Let $\theta$ be an angle in the interval $(0,\pi/2)$ . Given that $\cos \theta$ is irrational, and that $\cos p \theta$ and $\cos q \theta$ are both rational for some relatively prime positive integers $p, q$ , show that $\theta = \pi/6$ .

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#10 h Jan 4, 2023, 4:57 PM • 1 Y

Y by cubres

Nice problem

The difficulty of the problem lies in proving that $\theta$ is a rational multiple of $\pi$ . Suppose otherwise, and let $z=\operatorname{cis}(\theta)$ . Then since $\cos k\theta:=q_1$ is rational, we have
$\operatorname{Re}(z^k)=q_1 \implies z^k+z^{-k}=2q_1 \implies z^{2k}-2q_1z^k+1=0,$ so $z$ is a root of $P_1(x):=x^{2k}-2q_1x^k+1$ . Likewise, $z$ is a root of $P_2(x):=x^{2k+2}-2q_2x^{k+1}+1$ , where $\cos (k+1)\theta=q_2$ , so it's a root of $\gcd(P_1(x),P_2(x)) \in \mathbb{Q}[x]$ .
It turns out that we can find all roots of $P_1(x)$ without great difficulty: observe that all numbers of the form $\operatorname{cis}(\pm\theta+\tfrac{2n\pi}{k})$ are roots, where $n=0,\ldots,2k-1$ . Since there are $2n$ of these (all such roots are distinct because $\theta$ isn't a rational multiple of $\pi$ ), they are actually precisely the roots of $P_1$ . Likewise, the roots of $P_2$ are precisely $\operatorname{cis}(\pm \theta+\frac{2n\pi}{k+1})$ . Because $\theta$ isn't a rational multiple of $\pi$ and $k \perp k+1$ , the only shared roots between $P_1$ and $P_2$ are $\operatorname{cis}(\pm\theta)=z,\overline{z}$ . Thus
$\gcd(P_1(x),P_2(x))=(x-z)(x-\overline{z}) \in \mathbb{Q}[x],$ which implies that $z+\overline{z} \in \mathbb{Q} \implies \operatorname{Re}(z)=\cos \theta \in \mathbb{Q}$ : illegal. Thus $\theta$ must be a rational multiple of $\pi$ .

By Niven's Theorem, if $\phi$ is a rational multiple of $\pi$ and $\cos(\phi) \in \mathbb{Q}$ , then $\phi \in \{0,\pi/3,\pi/2,2\pi/3\} \pmod{\pi}$ , so modulo $\pi$ the only possible differences between two rational multiples of $\pi$ whose cosines are rational are $0,\pi/6,\pi/3,\pi/2,2\pi/3$ . Since the cosines of $k\theta$ and $(k+1)\theta$ are both rational, it follows that
$(k+1)\theta-k\theta=\theta \in \{0,\pi/6,\pi/3,\pi/2,2\pi/3\} \pmod{\pi},$ so if $\theta \in (0,\pi/2)$ then we must have $\theta \in \{\pi/6,\pi/3\}$ . But $\cos \pi/3$ is rational, hence $\theta=\pi/6$ as desired. $\blacksquare$

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#11 h Jan 15, 2025, 1:34 AM • 1 Y

Y by cubres

Here we go:

If $\theta$ is a rational multiple of $\pi$ , by Niven's theorem, $k \theta , (k+1) \theta \in \{0,\pi/3,\pi/2,2\pi/3\} \pmod{\pi}$ Thus, their difference $(k+1) \theta - k \theta = \theta \in \{0,\pi/6,\pi/3,\pi/2,2\pi/3\} \pmod \pi$ . Since $\theta \in (0, \pi/2)$ , and $\cos \theta$ is irrational, $\theta = \pi/6$ .

If $\theta$ is not a rational multiple of $\pi$ : Consider the following two polynomials: $P(x)= x^{2k}-2 \cos (k \theta) x^k+1, G(x) = x^{2(k+1)}-2 \cos ((k+1) \theta) x^{k+1}+1.$ Their roots are respectively: $z = \text{cis}\left( \pm \theta + \frac{2\pi s}{k} \right)$ $z = \text{cis} \left( \pm \theta + \frac{2\pi t}{k+1} \right).$ Since $\theta$ is not a rational multiple of $\pi$ : the only common roots are $\text{cis} ( \pm \theta )$ . Thus: $H(x) = \gcd(P, G) = (x-\alpha)(x-\bar{\alpha})$ where $\alpha = \text{cis}(\theta)$ . Since $P, G \in \mathbb Q [x ]$ , we have: $H \in \mathbb Q [ x ]$ . Thus: $\alpha + \bar{\alpha} = 2 \cos \theta \in \mathbb Q$ which contradicts that $\cos \theta$ is irrational.

This post has been edited 1 time. Last edited by Saucepan_man02, Jan 15, 2025, 1:35 AM
Reason: EDIT

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#12 h Mar 4, 2025, 9:39 PM • 1 Y

Y by cubres

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Please contact westskigamer@gmail.com if there is an error with any of my solution for cash bounties by 3/18/2025.

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#13 h Apr 12, 2025, 8:14 PM • 1 Y

Y by cubres

Assume $\theta$ wasn’t a rational multiple of $\pi.$ Denote $z = \cos\theta + i\sin\theta$
Consider the polynomials $P(X) = X^{2k} - 2\cos(k\theta) X^k + 1$ and $Q(X) = X^{2k + 2} - 2\cos[(k + 1)\theta]X^{k + 1} + 1.$
Note that $P$ has zeroes when $X^k = e^{\pm \pi i k\theta}$ and $Q$ has zeroes when $X^{k + 1} = e^{\pm \pi i(k+1)\theta}.$
It is obvious that they only share the roots $z, z^{-1}.$
Hence, $\gcd(P, Q) = (x - z)(x - z^{-1}) = x^2 - 2\cos\theta x + 1.$ However, note that $P, Q \in \mathbb Q[x]$ from the given constraints. Therefore, the gcd, $x^2 - 2\cos\theta x + 1$ must also be in $\mathbb Q[x].$ This implies that $\cos\theta\in\mathbb Q,$ which gives us a contradiction.

Thus, $\theta$ is a rational multiple of $\pi.$ By Nixen, it follows $\cos(k\theta) = \{-1,-1/2,0,1/2,1\}.$ Obviously, this implies $\cos\theta = \frac{\sqrt3}{2},$ or $\theta = \frac\pi6.$

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