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Why are trig-functions standard in radians as opposed to degrees?

I was wondering about this because when differentiating a trig-function in degrees one should apply the chain-rule: $$\frac d{dx}\sin x^\circ=\frac\pi{180}\cos x^\circ$$ Why could the degrees not be the standard?

Parcly Taxel
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GambitSquared
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    Exactly for the reason that we don't want factors of $\frac \pi{180}$ all over the place when doing calculus, I expect. –  Oct 03 '16 at 15:52
  • But I mean, why not have degrees without extra term when differentiating and have the extra term when differentiating in radians? – GambitSquared Oct 03 '16 at 15:57
  • That factor has nothing to do with radian measure. Had we never discovered radians, the derivative of the degree sine function would still be $\frac {\pi}{180}\cos(x)$. –  Oct 03 '16 at 15:58
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    This question has been asked... about 360 times here - it was asked yesterday http://math.stackexchange.com/questions/1951021/why-in-calculus-the-angles-are-measured-in-radians/1951068 Please try the search box before asking – leonbloy Oct 03 '16 at 15:58
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    because $\mathop {\lim }\limits_{x, \to ,0} \frac{{\sin x}}{x} = 1$ only when $x$ is in radians. – G Cab Oct 03 '16 at 16:01
  • "why not have degrees without extra term when differentiating and have the extra term when differentiating in radians?" Suppose we measured angles in kloopfs. then arc length of angle along a unit circle = circumference x kloopfs x scaling factor = 2pi x kloopfs x scaling factor. where scaling factor = #kloopfs in circle/2pi. Then we'd have sin' x = scaling factor x cos x. It's be nice if scaling factor = 1. That means # kloopfs in circle/2pi = 1. So # kloopfs in circle = 2pi. In other words... kloopfs are radians. – fleablood Oct 03 '16 at 16:18
  • In other words, we can't just declare a scaling factor to be 1 and ignore it. A scaling factor, k, must always exist and it must be $\frac {2\pi}{\text{# of units in the angle measure of a circle}}$. If we use 360 degrees in circle then $k = \frac{\pi}{180}$ If we use 400 metric degrees in a circle, $k =\frac{\pi}{200}$ and so on. We don't declare sin' x = cos x regardless of unit and require the conversion to go there based on an arbitrary unit. We must instead find our our arbitrary unit and discover sin' x = k cos x. where k is based on how our arbitrary unit relates to $2\pi$. – fleablood Oct 03 '16 at 16:35
  • Why do American's use inches and feet and all that when using meters and similar units are $\large\text{so}$ much easier?! – Simply Beautiful Art Oct 03 '16 at 23:18

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Degrees have no mathematical basis, actually it is not known with certainty where the measure originates from (though apparently it may be related to the fact that a year has approx. 360 days, meaning the earth rotates around the sun by ~1deg each day).

Radians however, make perfect mathematical sense. They represent the length of the arc delimited by the corresponding angle on a radius 1 circle.

It also turns out this makes most computations much more easy, in particular with complex numbers : the cosine and sine functions (and other trig functions) have simpler expressions in terms of the exponential function. Their Taylor expansions don't have pi's all over the place. Their derivatives are easy to express as a function of each other ; and the list goes on.

Ulysse Mizrahi
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