What is the difference between the derivative of a sine function between the degrees and radian values?

Question

Is the derivative of the sine function, where the angle is measured in degrees, the same as the derivative of the sine function, where the angle is measured in radians.?

Please spare me the mathematical equations i know them, i want a theoretical/intuitive explanation.

Answer 1

Define $S(x)$ and $C(x)$ to be the sine and cosine of the angle $x$ measured in degrees. Then $S(x) = \sin( \frac{\pi}{180} x)$ and $C(x) = \cos( \frac{\pi}{180} x)$ so that $$S'(x) = \frac{\pi}{180} \cos \left( \frac{\pi}{180} x \right) = \frac{\pi}{180} C(x).$$ Similarly, $C'(x) = - \dfrac{\pi}{180} S(x)$.

Answer 2

$\frac{d}{d\theta}[\sin(\theta)]=\cos(\theta)$ works only if $\theta$ is in radians.

If $\theta$ is in degrees and $t$ is $\theta$ converted to radians (i.e. if $t=\frac{180}{\pi}\theta),$ then: $$\underbrace{\color{green}{\frac{d}{d\theta}[\sin(t)]}=\cos(t) \cdot\frac{dt}{d\theta}}_{\text{chain rule}}=\boxed{\color{green}{\left[\frac{180}{\pi} \right] \cos(t)}}.$$

Answer 3

Here is an attempt at an intuitive explanation ...

If you apply a scale factor like this to $y=f(x)$, it is like stretching the $x$-axis, which (for stretch factors greater than $1$) flattens the curve, and hence the tangent to the curve.

Since the slope of the chord between $(x_1, y_1)$ and $(x_2, y_2)$ is $\cfrac {y_2-y_1}{x_2-x_1}$ the slope of the chord is changed by the stretch factor: the $y_i$ stay the same and the $x_i$ change. The same is true of the tangents, which are limits of the chords.