# What is there in the derivation of the Taylor/Maclaurin series for sin(x) that determines if the series assumes x is radians or degrees?

Apr 12, 2015

If the angle is not measured in radians, then the differentiation rule $\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$ is false.

In establishing (proving) that rule we use ${\lim}_{h \rightarrow 0} \frac{\sinh}{h} = 1$

That is also not true if $h$ is taken to be the measure of an angle in units other than radians.

And that's because proving ${\lim}_{h \rightarrow 0} \frac{\sinh}{h} = 1$ uses "arc length = central angle in radians times radius"

If the central angle $\theta$ is measured in degrees, the arc length, $s$ may be found by using the proportion:

$\frac{s}{C} = \frac{\theta}{360}$ or $\frac{s}{2 \pi r} = \frac{\theta}{360}$

With $\theta$ measured in radians, the $360$ is replaced by $2 \pi$.

If $\theta$ measures $m$ in degrees then measures $m \frac{\pi}{180}$ radians. Instead of the simple formulas:

$s = r \theta$ and $\frac{d}{d \theta} \left(\sin \theta\right) = \cos \theta$

We have:

$s = \frac{\pi r m}{180}$ and the derivative is $\frac{\pi}{180} \cos m$ or maybe it's $\frac{180}{\pi} \cos m$. (My brain hurts when I try to use degrees to do calculus.)

Apr 12, 2015

You could also just say that the idea of the MacLaurin series for $\sin \left(x\right)$ is so you can say that $\frac{d}{\mathrm{dx}} \left(\sin \left(x\right)\right) = \cos \left(x\right) = 1$ and $\sin \left(x\right)$ is approximately equal to $x$ near $x = 0$.

When you do $\sin \left(0.02\right)$, in radians it does return you an answer of 0.02 with negligible error (you get 0.01999867). When you get that, you know you're in the right units. In degrees, you get 0.9110879, which is not a 1:1 relation of the argument with the answer.

But this is just a quick check. The actual proof might not be like this; I don't remember how it actually went.