Why finding the derivative of trig functions, why must the angle be in radians?why not degrees?
It comes from the derivation of the derivative of sin x.
Let
 = \sin{(x)} )
Then
 - f(x) = \sin{(x+h)} - \sin{(x)} = 2\sin{(h/2)}\cos{(x + h/2)} )
(I won't prove this identity; uses the fact that sin(a+b) - sin(a-b) = 2 sin(b) cos (a), and then let x + h = a + b, x = a - b, solve for a and b and plug into above formula)
The derivative limit definition then is
 = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h\rightarrow 0} \frac{2\sin{(h/2)}\cos{(x + h/2)}}{h} )
The cosine term can be moved out by continuity of the cosine function. This gives
 = \cos{(x)} \lim_{h\rightarrow 0} \frac{\sin{(h/2)}}{h/2} )
It can be proved, using a geometric argument comparing the area of an inscribed triangle with the corresponding sector and a tangential triangle, that the final limit is 1 if and only if h is measured in radians. If it's measured in degrees, you'll need a conversion factor here.