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?
 A: 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,
$$
(\cos(\pi),\sin(\pi))=(-1,0)\tag{7}
$$
$\hspace{8mm}$
A: If you have a unit circle, a circle with radius 1, you can define the circumference in terms of $\pi$ and the radius using the formula $C=2\pi r$. Now here is a unit circle:
Using our formula we can deduce that the circumferance is $2\pi$. Now what this means is that if a person were to travel around the circle exactly on the line they would of traveled a distance of $2\pi$ in terms of $r$.
Now radians is a way to measure angles that has some nice properties that makes it much easier to work with in Mathematics than other angle systems such as degrees. 
$1$ radian is defined as the angle corresponding to an arc length of $1$ around a unit circle as shown.

Now if you remember we had deduced that the circumference (all the way around the outside) had a length of $2\pi$. So using our radian definition it means that the angle going all the way round in a circle must be $2\pi$ radians.
This is where we start to find out that the two ARE ACTUALLY RELATED!!
Now as radians is the natural way to measure angles many forumlae such as the taylor expansion of $\sin x$ are unaltered by conversion to other units of angles because it IS the system to use. Now look back to what $\sin$ and $\cos$ actually tell use about a circle. When we have an angle $\theta$ on a unit circle, the trigonometric functions of $\theta$ tell use about different lines on the unit circle. Using this angle we can construct a trianlge (hence the name trigonometry, measure of triangles).
Now we can see why we use radians as it appears that the trigonometric functions are relating the circumference and the radius to the angle $\theta$ and radians is the angle system that also does relating between the angle $\theta$ and the circumference.
A: The facts you mention are definitely not a coincidence; they are a consequence of how things are defined, and partly a reason why they are defined so. (This is an incredibly vague statement, but the question itself is a little vague).
Some textbooks will define $\sin x$ by the series expansions you just mentioned, and then define $\pi$ as the least such positive number that $\sin \pi = 0$. Also, put $\cos x = \sin' x$. You can then check that $(\cos x, \sin x)$ parametrize points on the circle, and then calculate the circumference of a unit circle using the appropriate integral. This is a legitimate way of doing mathematics, but not a very illuminating one (in this particular situation), I'm afraid. 
If you know a little about differential equations (and are prepared not to be absolutely rigorous), you can proceed as follows. Let's agree that you know that there are functions $\sin x $ and $\cos x$ (as functions of the angle, so that the precise units are yet to be determined). Denote the measure of the full angle by $\tau$ (there $\tau$ is just a name just now, nothing more), and the circumference of the unit circle by $C$. Note that we could set $\tau$ to be pretty much any number (like the well know constant $2 \pi$, or $360$, or whatever), but $C$ is fixed once and for all, and in fact $C = 2 \pi$.
Now, $\sin x$ and $\cos x$ will have some expansions as Taylor series, which will depend on what choice of $\tau$ we take (e.g. you have $\sin \tau/4 = 1$, but what this means depends on what $\tau$ is). We would like to choose $\tau$ so that the Taylor series for $\sin$ is something ``nice'' - which is of course a very vague requirement. To figure out the Taylor series, it's easiest to start with some differential equations. So let's try to find out what $\sin'x$ is. After some thought (it's probably best to draw a picture...) you can convince yourself that $\sin'x = \frac{C}{\tau} \cos x$. Likewise,  $\cos'x = - \frac{C}{\tau} \sin x$. Hopefully, it is clear that things will become nicest when $C = \tau$, so that $\sin' x = \cos x $ and $\cos'x = - \sin x$ - and this is the choice that the mathematicians made! 
With a little work, you can derive from the conditions $\sin' x = \cos x $ and $\cos'x = - \sin x$ the Taylor expansion for $\sin x $ and $\cos x$. You could set other units (i.e. choose different $\tau$) - only then the Taylor expansion would not look so nice.
A: I'm not sure why you started with the Taylor series as it does not seem relevant to your question.
To clarify:
$\pi$ is defined as the ratio of the circumference of a circle to its diameter.
$$\pi = \frac{circumference}{diameter}$$ 
and the radian is defined as the angle subtended by an arc on the circumference of a circle with length equal to the radius.
There are of course other measurements of angle such as the degree defined as the angle subtended by $\frac{1}{360}$ of the circumference of a circle.  Its thought that 360 was chosen because it can easily be divided by 2, 3, 4, 5, 6, 8, 9, 10 and 12.
Another measure of angle is the grad though it's use is less common than the other two. 
