Why do we require radians in calculus?
Radians make it possible to relate a linear measure and an angle measure. A unit circle is a circle whose radius is one unit. The one unit radius is the same as one unit along the circumference. Wrap a number line counter-clockwise around a unit circle starting with zero at (1, 0). The length of the arc subtended by the central angle becomes the radian measure of the angle.
From Why Radians? | Teaching Calculus
We are therefore comparing like with like the length of a radius and and the length of an arc subtended by an angle $L = R \cdot \theta$ where $L$ is the arc length, $R$ is the radius and $\theta$ is the angle measured in radians.
We could of course do calculus in degrees but we would have to introduce awkward scaling factors.
The degree has no direct link to a circle but was chosen arbitrarily as a unit to measure angles: Presumably its $360^o$ because 360 divides nicely by a lot of numbers.
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.
It really comes down to the following limit: $$ \lim_{x\to 0} \frac{\sin(x)}{x} = 1 $$ Or in other words, "$\sin x \approx x$ for small $x$". As a consequence, we have $$ \frac{d}{dx}\sin x = \cos x, \qquad \frac{d}{dx}\cos x = -\sin x $$ For any other choice of angular unit, these derivatives require some sort of coefficient (such as $\pi/180$). In this sense, radians are the "natural" unit for an angle, as far as calculus is concerned.