Rational values

There exists a rational number k such that $cos(k\pi)=sin(x\pi)$ . Denote $k=\frac {p}{q},cos(k\pi)=a \in Q$ There exists one and only polynomial $f_n(x) \in Z[x]$ satisfies $f_n(2cos\theta)=2cos(n\theta), \forall \theta \in R$ . $deg f_n(x)=n$ and the coefficient of highest degree =1. This can easily be proved by induction as $f_1(x)=x,f_2(x)=x^2-2$ and $f_{n+1}(x)=xf_n(x)-f_{n-1}(x)$ hence $f_q(2a)=f_q(2cos(\frac {p}{q} \pi))=2cos(p\pi)=2(-1)^p$ Then 2a is a rational root of polynomial with integer coefficient $f_q(2a)-2(-1)^p$ the coefficient of highest degree =1 $\implies$ $2a \in Z$ . $|a| \leq 1$ $\implies$ $|2a| \leq 2$ $\implies$ $2a \in {0,1,-1,2,-2}$ $\implies$ $a \in \{0,\frac {1}{2},\frac {-1}{2},1,-1\}$ $sin(x\pi)= a$ then it is easy to check that there are nine values of x satisfy.

Solution 1: We will first consider the problem for $\cos x\pi,$ instead of $\sin x\pi$ and prove that if $x$ and $\cos x\pi$ are both rational, then $\cos x\pi$ is $0,$ $\pm \frac{1}{2},$ or $\pm 1.$ To do this, consider a sequence of rational numbers $a_n=2\cdot \cos (2^n x\pi),$ $n=0,1,2,...$ Note that for all $n\geq 1$ $a_n=a_{n-1}^2-2.$ So if $a_0$ is a rational number with the denominator (in simplest terms) $q>1$ , then $a_1$ is a rational number with the denominator $q^2,$ $a_2$ has denominator $q^4$ and so on. Since $x$ is rational, all multiples of $x\pi$ modulo $2\pi$ belong to a finite set, so the numbers $a_n$ belong to a finite set, and their denominators are bounded. This is a contradiction, so $a_0=2\cos x\pi$ must be an integer.

Since $\sin x\pi=\cos ((\frac{1}{2}-x)\pi),$ we conclude that $\sin x\pi$ for a rational $x$ must be one of the numbers $-1,\ -\frac{1}{2},\ 0,\ \frac{1}{2},\ 1.$ We can count that there are $1 + 2 + 3 + 2 + 1 = 9$ solutions.

Solution 2: Since $\left( e^{\pm i\pi \frac{p}{q} } \right)^ q - (-1)^p = 0$ ,hence $e^{\pm i \pi \frac{p}{q} }$ is an algebraic integer. Thus, $2 \sin \left( \frac{p}{q} \pi \right) = -i\left(e^{i \pi \frac{p}{q} } - e^{-i \pi \frac{p}{q} }\right)$ , which is the difference and product of algebraic integers, is also an algebraic integer.

Suppose we have $\frac{p}{q}$ such that $\sin \left(\frac{p}{q} \pi \right)$ is rational, then $2\sin \left( \frac{p}{q} \pi \right)$ is a rational algebraic integer. However, the only rational algebraic integers are the integers themselves, which means that $2\sin \left( \frac{p}{q} \pi \right)$ is an integer. Since $-2 \leq 2 \sin \theta \leq 2$ , hence this implies that we must have $\sin \left( \frac{p}{q} \pi\right) \in \{ -1, -\frac{1}{2}, 0, \frac{1}{2}, 1 \}$ .

We can count that there are $1 + 2 + 3 + 2 + 1 = 9$ solutions.

Note: This is known as Niven's Theorem .

Niven's theorem states that the only rational values of θ in the interval 0 ≤ θ ≤ $\frac{\pi}{2}$ for which the sine of θ degrees is also a rational number are:

$\sin 0 = 0$ on the x axis

$\sin \frac{\pi}{6} = \frac 12$ in between x and y axis

$\sin \frac{\pi}{2} = 1$ on the y axis

Thus, there are four rational values along the x and y axis and four in between the axis for 0 ≤ x < 2. The ninth rational value is for x = 2.

Niven’s Theorem states that if $\frac {r} {\pi}$ and $\sin r$ are both rational, then the sine take values 0, $\pm \frac {1} {2}$ , and $\pm 1$ .

Let $x= \frac {r} {\pi}$

Based on Niven’s Theorem that states if $x$ and $sin (\pi x)$ are both rational, then the sine take values 0, $\pm \frac {1} {2}$ , and $\pm 1$ .

Therefore, there are $9$ values of $x$ , which are $0, \frac {1} {6}, \frac {1}{2}, \frac {5} {6}, 1, \frac {7} {6}, \frac {3} {2}, \frac {11} {6}, 2$ producing the sine values of $0, \frac {1}{2}, 1, \frac {1} {2}, 0, -\frac {1} {2}, -1 -\frac{1} {2},0$ respectively.

The interval for $x$ , $[0,2]$ , translates into the interval $[0, 2\pi]$ for $x\pi$ , or one full period for a sine function.

From one full period, the obvious rational values are the maxima, minima, and infection points, $0$ , $1$ , and $-1$ . This yields a total of $\fbox{5}$ values.

For the values in between maxima/minima and inflection points, the only rational one is $\frac{1}{2}$ , given by Niven's Theorem . This yields $\fbox{4}$ more values, one for each quadrant.

$4 + 5 = 9$ giving us $\fbox{9}$ total rational values.

Using Niven's Theorem, $\text{ if } y \text{ and } \sin \pi y \text{ are rational, }\sin x=0,\pm 1,\pm \frac12$

$\sin x=0, x=n\pi\text{ as } 0\le n\pi\le 2\pi\implies 0\le n\le 2$

$\sin x=\pm 1\implies \cos x=0, x=n\pi+\frac\pi2 \text{ as } 0\le n\pi+\frac\pi2\le 2\pi , 0\le n\le 1$

$\sin x=\pm \frac12\implies \sin^2x=\frac14=\sin^2\frac\pi6\implies x=n\pi\pm \frac\pi6,$

$\text{If }x=n\pi+\frac\pi6, \text{ as } 0\le n\pi+\frac\pi6\le 2\pi \implies 0\le n\le1$

$\text{If }x=n\pi-\frac\pi6, \text{ as } 0\le n\pi-\frac\pi6\le 2\pi \implies 1\le n\le2$

sin(pix) has only the following rational values at rational x:- 0, ±1/2, ±1 the corresponding values of x are : 0,1/6,1/2,5/6,1,7/6,3/2,11/6,2 Hence, there are 9 possible values of x.

From the Niven's Theorem (or the unit circle if you don't like being rigorous), we know that the only values of $0\le x\pi\le2\pi$ such that $x\in\mathbb{Q}$ and $\sin(x\pi)\in\mathbb{Q}$ are

$\sin{0}=\sin{\pi}=\sin{2\pi}=0$ ,

$\sin{\frac{\pi}{2}}=1$ ,

$\sin{\frac{3\pi}{2}}=-1$ ,

$\sin{\frac{\pi}{6}}=\sin{\frac{5\pi}{6}}=\frac{1}{2}$ , and

$\sin{\frac{7\pi}{6}}=\sin{\frac{11\pi}{6}}=-\frac{1}{2}$

Thus, there are 7 values of $x$ .

Niven's theorem states that the only rational values of $\theta$ in the interval 0 ≤ θ ≤ 90 for which the sine of θ degrees is also a rational number are:

$\sin 0^o = 0; \sin 30^o = \frac{1}{2}; \sin 90^o = 1.$

if θ is in whole degrees , then θ° = πx/180 radians = π*(p/q) radians

so we wish to find all rational multiples of π so that sin((p/q)*π)) rational

is p/q ϵℚ then e^( ±iπ*(p/q)) is an algebraic integers since

(e^( ±iπ*(p/q))^q-(-1)^p = 0

thus

2 sin(π (p/q)) = -i(e^(iπ (p/q))-e^(-iπ*(p/q)) is the difference and product of

algebraic integers, and therefore an algebraic integer.

However, the only rational algebraic integers are normal integers

Thus, the only values of sin((p/q)*π) which could be rational,

are those for which 2 sin(π*(p/q))

is an integer,then that imply sin(π*(p/q)) ϵ {-1,-1/2,0,1/2,1}

sin (π*x) = y where x,y are rationals then y must be 0±1/2±1

therefore we see that sin (π*x) rationals on interval x {0,2}

0,π/6,π/2, 5π/6, π, 7π/6, 3π/2, 11π/6, 2π

Niven's theorem states that the only rational values of , such that is rational are .

Now . So is also rational for .

Using cyclotomic fields it can be proved that is rational only if . Since is rational and and are rational.

Therefore we see that is only rational 0, pi/6, pi/2, 5pi/6, 1pi, 7pi/6, 3\pi/2, 11pi/6, 2pi