Define two rational numbers $\alpha$ and $x$ such that $\sin( { \alpha }) =x$

Question

Of course for $x\neq 0 $ and $\alpha$ in radians. Can you define them?

Answer 1

When $\alpha$ is rational and nonzero, $\sin \alpha$ is transcendental. This follows from the Lindenmann-Weierstrass theorem.

Answer 2

I think the Lindenmann-Weierstrass theorem is an overkill for this problem. Ivan Niven got an elementary proof of

For any rational $r\neq 0$, $\cos r$ is irrational.

in his book Irrational Numbers by exploiting the fact that rational number being the roots of a polynomial, the proof is not lengthy, and is very inspiring, I learned the proof back in college days. Here I just copy the proof (thus making the answer a community wiki) and adapt it for sine function:

Claim: For any rational $r\neq 0$, $\sin r$ is irrational.

For a positive $r = a/b$, where $a,b\in \mathbb{Z}^+$. For $p$ being an odd prime to be specified, define $$ f(x) = \frac{x^{p-1}(a-bx)^{2p}(2a-bx)^{p-1}}{(p-1)!} = \frac{(r-x)^{2p}\big(r^2 - (r-x)^2\big)^{p-1}b^{3p-1}}{(p-1)!}\tag{1} $$ For $0<x<r$: $$ 0<f(x)< \frac{r^{2p}(r^2)^{p-1}b^{3p-1}}{(p-1)!} = \frac{r^{4p-2}b^{3p-1}}{(p-1)!} \tag{2} $$ Next, define $F(x)$ using all the even derivatives of $f(x)$ $$ F(x) = f(x) - f''(x) + f^{(4)}(x) - f^{(6)}(x) + \cdots - f^{(4p-2)}(x). $$ Thus $$ \big(F'(x)\cos x + F(x)\sin x\big)' = F''(x)\cos x + F(x)\cos x = f(x)\cos x $$ and $$ \int^r_0 f(x)\cos x \,dx = F'(r)\cos r + F(r)\sin r + F(0). \tag{3} $$ Next, what Ivan Niven does is proving $F'(r)=0$ by construction to get rid of $\cos$, computing $F(r)$ and $F(0)$, setting $\sin r$ to be rational, then to get a contradiction.

• $F'(r)=0$ follows from $f(x)$ has a factor of $(r-x)^2$.

• For $f(x)$ has form $x^{p-1}g(x)/(p-1)!$, $g(x)$ has all integer coefficients, $f^{(j)}(0)$ is an integer, also notice $f^{(j)}(0)$ is a multiple of $p$ unless $j=p-1$. Direct computation from (1) gives $$ f^{(p-1)}(0) = a^{2p}(2a)^{p-1}. $$ If $p>a$, and $p$ is an odd prime, then $f^{(p-1)}(0)$ is not divisible by $p$. Thus $F(0) = q$, where $q$ and $p$ are relatively prime.

• The definition (1) implies that $$ f(r-x) = \frac{x^{2p}(a^2 - b^2x^2)^{p-1}b^{p+1}}{(p-1)!}. $$ Using similar argument as above, $f^{(j)}(r)$ is a multiple of $p$. Hence let $F(r) = pm$ for some integer $m$.

Now assume

$\sin r = d/k$ is rational for $d,k\in \mathbb{Z}^+$.

It suffices to show for positive sine value for $\sin(\cdot)$ is an odd function. Plugging all above results into (3): $$ k\int^r_0 f(x)\cos x \,dx = pmd + kq\tag{4} $$ The contradiction will be obtained by choosing $p$:

Choose $p$ large enough such that the right side of (4) is a non-zero integer, and left side is strictly in $(-1,1)$.

For $p>a$ is already required, now let $p>k$, thus $p$ is relatively prime with $kq$, thus right side of (4) is a non-zero integer. Now by (2): $$ \left|k\int^r_0 f(x)\cos x \,dx\right| < kr^3b^2\frac{(r^4b^3)^{p-1}}{(p-1)!} $$ Since $$ \lim_{p\to \infty} \frac{(r^4b^3)^{p-1}}{(p-1)!} = 0 $$ So we can choose $p$ large enough so that the left side of (4) is strictly in $(-1,1)$. Contradiction, hence the claim holds.

---

Notice Ivan Niven's proof tells us nothing about whether $\sin r$ can be algebraic. For that part I believe we have to borrow the Lindenmann-Weierstrass theorem, if I am wrong here someone please point out.