Why do we require radians in calculus? I think this is just something I've grown used to but can't remember any proof.
When differentiating and integrating with trigonometric functions, we require angles to be taken in radians. Why does it work then and only then?

To make commenters' points explicit, the "degrees-mode trig functions" functions $\cos^\circ$ and $\sin^\circ$ satisfy the awkward identities
$$
(\cos^\circ)' = -\frac{\pi}{180} \sin^\circ,\qquad
(\sin^\circ)' = \frac{\pi}{180} \cos^\circ,
$$
with all that implies about every formula involving the derivative or antiderivative of a trig function (reduction formulas for the integral of a power of a trig function, power series representations, etc., etc.).

Added: Regarding Yves Daoust's comment, I read the question, "Why does it work [if angles are taken in radians] and only then?", as asking, "Why do the derivative formulas for $\sin$ and $\cos$ take their familiar form when (and only when) $\sin$ and $\cos$ are $2\pi$-periodic (rather than $360$-periodic)?" If this interpretation is correct, and if one accepts that one full turn of a circle is both $360$ units of one type (degrees) and $2\pi$ of another (radians), then the above formulas are equivalent to $\sin' = \cos$ and $\cos' = -\sin$, and (I believe) do justify "why" we use the $2\pi$-periodic functions $\cos$ and $\sin$ in calculus rather than $\cos^\circ$ and $\sin^\circ$.
Of course, it's possible naslundx was asking "why" in a deeper sense, i.e., for precise definitions of "cosine and sine in radians mode" and a proof that $\cos' = -\sin$ and $\sin' = \cos$ for these functions. 
To address this possibility: In my view, it's most convenient to define cosine and sine analytically (i.e., not to define them geometrically), as solutions of the second-order initial-value problems
\begin{align*}
\cos'' + \cos &= 0 & \cos 0 &= 1 & \cos' 0 = 0, \\
\sin'' + \sin &= 0 & \sin 0 &= 0 & \sin' 0 = 1.
\end{align*}
(To say the least, not everyone shares this view!) From these ODEs, it's easy to establish the characterization:
$$
y'' + y = 0,\quad y(0) = a,\ y'(0) = b\quad\text{iff}\quad
y = a\cos + b\sin.
$$
One quickly gets $\cos' = -\sin$ and $\sin' = \cos$, the angle-sum formulas, power series representations, and periodicity (obtaining an analytic definition of $\pi$). After this, it's trivial to see $\mathbf{x}(\theta) = (\cos \theta, \sin \theta)$ is a unit-speed parametrization of the unit circle (its velocity $\mathbf{x}'(\theta) = (\sin\theta, -\cos\theta)$ is obviously a unit vector). Consequently, $\theta$ may be viewed as defining a numerical measurement of "angle" coinciding with "arc length along the unit circle", and $2\pi$ units of this measure equals one full turn.
