Is it necessary to have $\theta$ in radians to obtain $\frac{\sin{\theta}}{\theta} \to 1$ as $\theta\to 0$? $$\lim_{\theta \to 0} \frac{\sin{\theta}}{\theta} = 1$$
Could anyone explain why $\theta$ is in radians?
Is it because if $\theta$ is not in radians then $\frac{\sin{\theta}}{\theta} \neq 1$?
Is there a detailed explanation?
 A: $$
\lim_{x\to0}\frac{\sin x^\circ}{x} = \lim_{x\to0}\frac{\sin_\text{(radians)}\left(\frac{\pi x}{180}\right)}{x}.
$$
Now let $u=\dfrac{\pi x}{180}$.  Notice that as $x\to0$, we have $u\to0$.  And $x=\dfrac{180u}{\pi}$.  So the limit becomes
$$
\lim_{u\to0}\frac{\sin_\text{(radians)}\left( u \right)}{ \left( \frac{180u}{\pi} \right) } = \lim_{u\to0} \frac{\pi}{180} \cdot \frac{\sin_\text{(radians)} u}{u} = \frac{\pi}{180} \cdot \lim_{u\to0} \frac{\sin_\text{(radians)} u}{u} = \frac{\pi}{180}\cdot 1= \frac{\pi}{180}.
$$
Therefore
$$
\lim_{x\to0}\frac{\sin x^\circ}{x} = \frac{\pi}{180}.
$$
Only when radians are used is the limit $1$.
Likewise, only when radians are used is $\sin'$ equal to $\cos$.  If degrees are used then $\sin' = \frac{\pi}{180}\cdot\cos$.
This is similar to the situation with derivatives of exponential functions:
$$
\frac{d}{dx} a^x = \left(a^x\cdot\text{constant}\right).
$$
Only when $a=e$ is the constant equal to $1$.  That's what's "natural" about $e$.
Later note: I'd add a graphic to back this up if I could conveniently draw it and upload it.  But here I'll describe it.  First look at this: http://upload.wikimedia.org/wikipedia/commons/4/45/Unitcircledefs.svg Then imagine what happens if $\theta$ is infinitely small but positive.  In that case, the curved arc you're looking at looks like a straight line, and the vertical line that defines the sine is the same straight line.  So the ratio of their lengths is $1$.  But if you use some units besides radians, so that the length of the arc is not measured in the same units as the length of the vertical line, then their lengths are two different numbers, even though their lengths are the same, so the ratio of those numbers is not $1$.
A: In the xy-plane, draw a circle with radius 1 around the origin.
Given a point $(x,y)$ on the circle, the angle between the positive $x$-axis and the vector from the origin to $(x,y)$ in radians is just the length of the part of the circle from the point $(1,0)$ counterclockwise to $(x,y)$. If the angle is small, then $x$ and the length of the part of the circle are almost the same.
This is my sloppy way of saying that $\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$.   This limit is only $1$ if $\theta$ is given in radians.
If the angle $\theta$ is given in degrees, then in radians this is 
$\frac{2\pi}{360}\theta$.  By the usual limit rules, in case $\theta$ is given in degrees, the limit is $\frac{\pi}{180}$.
A: As pointed out in one comment, everything depends on your definition of the sine function. If you fix one sine function, then nothing changes, since it is just a change of variables by dilation. But radians are pure numbers, while degrees are numbers with a unit, and we should probably remark that $90^\circ$ is not a real number. If we denote by $\sin$ the sine function defined on $\mathbb{R}$, the expression $\sin 90^\circ$ must be interpreted as
$$
\sin \left(\frac{90^\circ}{180^\circ} \pi \right).
$$
A: The length $L$ of an arc of a circle with radius $r$ and subentending angle $\theta $ is 
$$\begin{equation*}
L=r\theta 
\end{equation*}$$
if and only if $\theta $ is measured in radians$^1$. Under this assumption and for $0<\theta<\pi/2$ radians
$$\begin{equation*}
\sin \theta <\frac{L}{r}=\theta <\tan \theta 
\end{equation*}$$
and 
$$\begin{equation*}
1<\frac{\theta }{\sin \theta }<\frac{1}{\cos \theta },
\end{equation*}$$
which implies that
$$\begin{equation*}
\lim_{\theta \rightarrow 0}\frac{\theta }{\sin \theta }=1\Leftrightarrow
\lim_{\theta \rightarrow 0}\frac{\sin \theta }{\theta }=1,
\end{equation*}$$
because $\frac{\sin \theta }{\theta }$ is an even function.
$^1$If $\theta $ is measured in degrees, then
\begin{equation*}
L=\frac{\pi r}{180}\theta .
\end{equation*}
A: The apparent confusion stems from an overloading of $\sin$, the function: the function that gives the sine of an angle when measured in degrees is different from the function that gives the sine of an angle when measured in radians. They are related by a scale in the input and so the limit changes with that scale.
A: It doesn't matter whether theta is in radians/degrees/what-so-ever. The result will remains the same (as long as the the units you use are linear to radians. meaning, theta units could not be radians squared, because then the limit will change (and will be 0 in that case).
A: "Angular distance" is not a real number. However, they are closely related: given a particular choice of angular distance unit (e.g. "radian"), one can convert a real number $x$ into an angular distance by writing "$x$ radians". Conversely, given any angular distance, we can divide out the unit to obtain a real number.
"Angular displacement" is another related concept. Every angular distance gives an angular displacement. The difference is, e.g., that the displacements given by "0 degrees" and "360 degrees" are the same displacement, despite being different distances.
Now, the confusion you're observing comes from two different uses of the term $\sin$. Geometrically, we define $\sin$ as a function of angular displacement, e.g. by the usual "opposite / hypotenuse" definition and its variants. This can be directly viewed as a function of angular distances as well.
But don't we usually write $\sin$ as a function of real numbers? Well, if we have canonically chosen a unit of angular distance, then we can convert between real numbers and angular distance, and so we can define $\sin x$ as, say, $\sin(x\ \text{radians})$.
But the function so defined is really a function both of $x$ and the chosen unit of angular distance! If you change the unit, you change the function.
There are many good non-geometric reasons why "$\sin(x\ \text{radians})$ is a useful function of real numbers. So, today, mathematicians use the notation $\sin x$ to be that function, rather than the geometric one.
So if you want to use $\sin$ to find "opposite / hypotenuse", then you have to convert angular distance to radians to get the number to plug into $\sin$.
