How did Leibniz prove that $\sin (x)$ is not an algebraic function of $x$? In the Wikipedia article about transcendental numbers we can read the following: 
The name "transcendental" comes from Leibniz in his 1682 paper where he proved that sin(x) is not an algebraic function of x.
I would like to know can someone reproduce here the proof of Leibniz of the non-algebraicity of $\sin$ or point me to the place on the internet where we have this proof.
I do not know German or Latin so if someone has the link where this is proved it would be OK if it is written in English.
Although I am an amateur it seems to me that in the time of Leibniz the techniques for proving the non-algebraicity of functions were not developed enough so it would be nice to see how he proved that for $\sin$.
 A: Leibniz's paper with the term "transcendental" is : DE VERA PROPORTIONE CIRCULI AD QUADRATUM CIRCUMSCRIPTUM IN NUMERIS RATIONALIBUS EXPRESSA, Act.Erudit.Lips.1682.
Reprinted into:


*

*Gottfried Wilhelm von Leibniz, Leibnizens mathematische Schriften, herausgegeben von C.I.Gerhardt (1858), page 118-on; see page 120:



transcendens inter alia habetur per aequationes gradus indefiniti.

For the proof, see: PRAEFATIO OPUSCULI DE QUADRATURA CIRCULI ARITHMETICA, page 93-on; see page 98.
A: I don't know how Leibniz did it, but it seems to me that the main ingredients of this straightforward proof ought to have been available to him (if phrased with less rigor than we do today):
Definition. A function $f$ of one variable is algebraic if there is a polynomial $p(x,y)$ such that $p(x,f(x))=0$ for all $x$.
Claim. For every nonzero polynomial $p$, there is an $x_0$ such that $p(x_0,\sin x_0)\ne 0$.
Proof by long induction on the highest power of $y$ that appears in $p(x,y)$.
First suppose that $p$ contains at least one nonzero term that doesn't contain $y$. Then $p(x,\sin x)=p(x,0)$ whenever $x$ is a multiple of $\pi$. But the right-hand side is a nonzero polynomial in one variable, and can therefore have at most finitely many zeroes. So there exist some $k\pi$ where $p(k\pi,0)$ -- and therefore also $p(k\pi,\sin k\pi)$ -- is nonzero.
On the other hand, if every term of $p$ contains $y$, then $p(x,y)=y q(x,y)$ for some polynomial $q$. The induction hypothesis applies to $q$, so let $x_0$ be a value such that $q(x_0,\sin x_0)\ne 0$. If $x_0$ happens to be a multiple of $\pi$, then because $q(x,\sin x)$ is clearly continuous, there will be another suitable $x_0$ nearby. So without loss of generality $x_0$ is not a multiple of $\pi$, and therefore $q(x_0,\sin x_0)\ne 0$ implies that
$$ p(x_0,\sin x_0) = \sin x_0 \cdot q(x_0,\sin x_0)\ne 0 $$
as required.
(Note that this proof depends only on the fact that the sine function is continuous and has infinitely many isolated zeroes).
A: (1) $\sin x$ is not an algebraic function of $x$
(2) $\sin x$ is a transcendental function
Statement $1$ is false. Statement $2$ is true
Counterexample to Statement $1$: $x = 0$
$\sin 0$ is an algebraic function of $0$
$\sin 0 = 0 = 2(0)^2 + 1(0) + 0$
Right side of equation is an algebraic function, a single-variable quadratic polynomial with coefficients $2$, $1$, $0$
Zero is a member of the exceptional set of the transcendental function $\sin x$
Leibniz proved Statement $2$. He did not prove Statement $1$
A: If $\frac{e^{i x} - e^{-i x}}{2 i}$ were an algebraic function of $x$, then $e^x$ would too ( $x\mapsto ix$). However, $e^x$ grows too fast for that ( to get easily some bounds for roots of polynomials, use a companion matrix).
