<s>[INST] Trigonometric Coincidence I Know that using Taylor Series, the formula of $\sin x$ is
$$x-x^3/3!+x^5/5!-x^7/7!\cdots,$$
and the unit of $x$ is radian (where $\pi/2$ is right angle).
However, the ratio of the circumference and the diameter of a circle is also $\pi$. Is it a coincidence? Or is there a proof?
 [/INST] Let's start with your series and take a couple of derivatives:
$$
\begin{align}
v(x)&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\dots\\
u(x)=v'(x)&=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\dots\\
u'(x)=v''(x)&=\hphantom{1}-\,\,x\,\,\,+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-\dots\\[6pt]
&=-v(x)
\end{align}
$$
That is,
$$
\begin{align}
v'&=u\\
u'&=-v
\end{align}\tag{1}
$$
Furthermore,
$$
\begin{align}
u(0)&=1\\
v(0)&=0
\end{align}\tag{2}
$$
Consider the derivative of $u^2+v^2$ :
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(u^2+v^2\right)
&=2uu'+2vv'\\
&=-2uv+2uv\\[6pt]
&=0
\end{align}
$$
Therefore, $u^2+v^2$ is constant. Since $u(0)^2+v(0)^2=1$, we must have
$$
u^2+v^2=1\tag{3}
$$
which, by $(1)$, also says that
$$
u^{\prime\hspace{1pt}2}+v^{\prime\hspace{1pt}2}=1\tag{4}
$$
$(3)$ says that $(u,v)$ lies on the unit circle and $(4)$ says that $(u,v)$ moves at unit speed (where $x$ is time); that is, the length of the arc swept out below is $x$:
$\hspace{35mm}$
Thus, if we measure the angle at the center of a unit circle subtended by an arc as the length of that arc, we have the position of the point on the circle at angle $x$ is $(u(x),v(x))$. No coincidence.
These functions are more commonly called the sine and cosine of the angle $x$:
$$
\begin{align}
\cos(x)&=u(x)\\
\sin(x)&=v(x)
\end{align}\tag{5}
$$
The circumference of a unit circle is $2\pi$, so a quarter circle (subtending a right angle) is $\pi/2$. Thus,
$$
(\cos(\pi/2),\sin(\pi/2))=(0,1)\tag{6}
$$
A half circle (subtending a straight angle) is $\pi$; therefore,
$$
(\cos(\pi),\sin(\pi))=(-1,0)\tag{7}
$$
$\hspace{8mm}$
</s>