Why should we convert degree to radian while differentiating? While graphing the sine function, it is always assumed that x is in radians.Similarly while differentiating the sinx function x is always in radians and we have to multiply by a constant if x is given to us in degrees, before differentiating. Is radian just a convention for trigonometric functions or is there some other reason?Why are we able to differentiate a function directly when x is in radians and not when it is in degrees?
 A: Let $f(x)=\sin x$ radians, and $g(x)=\sin x$ degrees.
Then
$$f'(x)=\cos x\text{ radians}$$
and
$$g'(x)=\frac{\pi}{180}\cos x\text{ degrees}.$$
Which is a more convenient sine function: $f$ or $g$?
The slope of the graph of $f$ at the origin is $1$, that of $g$ is $\pi/180$.
When you get to Maclaurin series you find
$$f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$
and
$$g(x)=\frac{\pi x}{180}-\frac{\pi^3x^3}{180^33!}+\frac{\pi^5x^5}{180^55!}-\cdots.$$
A particle moving so that its at $(\sin t\text{ radians},\cos t\text{ radians})$ after $t$ seconds is rotating round the unit circle 
an one unit per second,
a particle moving so that its at $(\sin t\text{ degrees},\cos t\text{ degrees})$ after $t$ seconds is rotating round the unit circle 
at $\pi/180$ units per second.
Are you still keen on degrees?
A: You could do everything in degrees... but then the formulas would look considerably uglier.
Suppose I give you an angle $x$ degrees. Then it corresponds to $\pi x/180$ radians. Let's define $\sin^\circ(x)$ and $\cos^\circ(x)$ to be the sine and cosine of the angle $x^\circ$, in degrees. (We still let $\sin(x),\cos(x)$ be the sine and cosine of $x$ in radians.) Then
$$
\frac{d}{dx}\sin^\circ(x) = \frac{d}{dx}\sin\left(\frac{\pi x}{180}\right) = \frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right) = \frac{\pi}{180}\cos^\circ(x)
$$
Bleh. That factor of $\pi/180$ is really ugly and annoying. The fact that we get no ugly constant when doing calculus with radians is precisely the reason we use radians for trigonometry.
