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Up until now, I've been using rads and degrees interchangeably, simply using the $^{\circ}$ symbol to signify degrees, and then using the correct trigonometric function, so that: $$sin(90^\circ)=sin(π/2)$$

I would think that the same line of thought could be used when dealing with complex exponentials, since x appears to always end up in a trigonometric function: $$e^{ix}=cos(x) + isin(x)$$

However this seems to completely break down when logarithms are brought into the picture: $$ln(e^{ix})=i(x + 2κπ),\hspace{1em}κ\in\mathbb{Z}$$

But (assume $κ=0$ for simplicity's shake) $$ln(5e^{i90^\circ}) = ln5 + i90^\circ$$ Isn't the same number as $$ln(5e^{iπ/2}) = ln5 + iπ/2$$

My textbook (on electronic circuit analysis) tells me to use radians here, but there is no mention as to why. Any help would be greatly appreciated.

gtsiam
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    Please, be nice to me and forget degrees forever (at least when doing mathematics) ! Cheers. – Claude Leibovici Jun 25 '20 at 12:46
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    Writing $X^\circ$ is just another way of writing $X \cdot \frac{\pi}{180}$ radians. If you think about the degree sign $\circ$ as indicating multiplication by $\frac{\pi}{180}$, then the two numbers really are the same – Ben Grossmann Jun 25 '20 at 12:47
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    Well that's what I'm doing, but it's really bothering me: What makes radians so special here? It shouldn't matter, should it? It does for some reason though... – gtsiam Jun 25 '20 at 12:48
  • @ClaudeLeibovici Unfortunately, the use of degrees in this fashion is common, one cannot "forget them forever" – Ben Grossmann Jun 25 '20 at 12:48
  • @gtsiam it absolutely matters! Long story short, as soon as either Euler's formula or any sort of calculus is involved, it is necessary to make the assumption that the argument is in radians. – Ben Grossmann Jun 25 '20 at 12:49
  • @ClaudeLeibovici: my dream would be the existence of a "revolutions" unit, which would spare us a lot of $\pi$'s and $2\pi$'s. –  Jun 25 '20 at 12:49
  • @Omnomnomnom Thanks, I'll keep that in mind! EDIT: I mean, I get it matters, but I don't get why. It's just a dimensionless unit in the end, isn't it? – gtsiam Jun 25 '20 at 12:50
  • @gtsiam For more information, see this post and this post. – Ben Grossmann Jun 25 '20 at 12:50
  • consider the question of what ${\sqrt{\pi \over 2}}$ is in degrees – user619894 Jun 25 '20 at 12:50
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    I read about this on Quora, I don't know if someone posted on this site too https://www.quora.com/Both-radians-and-degrees-are-dimensionless-then-why-cant-we-write-e-180i-1/answer/Roman-Andronov?ch=3&share=32a8643d&srid=XvUng – UmbQbify Jun 25 '20 at 12:53
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    @YvesDaoust. What a beautiful dream ! Do you want we work the problem together ? I should dedicate the ramining of my life to this topic. Cheers. – Claude Leibovici Jun 25 '20 at 12:53

2 Answers2

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Why it matters:

In degrees, Euler's formula would read

$$e^{iz\pi/180}=\cos(z)+i\sin(z)$$ and the whole world would hate that $\frac\pi{180}$ factor.

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    So "in short, it's in euler's formula". Well that I can get behind, thanks! Seems so obvious in retrospect and yet... – gtsiam Jun 25 '20 at 12:55
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    @gtsiam: another, maybe more convincing way, is that the circumference of the unit circle is $2\pi$, not $360$. –  Jun 25 '20 at 12:58
  • I did some more research and found the missing link in my reasoning.

    I don't really wanna answer my own question after you've helped me so much, so maybe you can expand the answer a bit.

    To show why euler's formula would read what you wrote (Note: "->" means "left depends on right for proof"): Euler's identity -> Maclauren series for sine -> sine derivative -> limit sinx/x as x goes to 0 = 1. Turns out the aforementioned limit is only 1 if x is in degrees, because the proof for it uses the length of an arc (That ties to your previous comment as well).

     – gtsiam Jun 25 '20 at 13:45
  • @gtsiam: there is no unique road. Sometimes the sine/cosine are defined from the complex exponential and $\pi$ is defined to be the half-period of these functions, hence the natural angle unit. –  Jun 25 '20 at 14:14
  • @Yvez Daoust: Fair enough, but for me this us what made thus click. – gtsiam Jun 25 '20 at 15:12
  • Actually just saying that you can prove Euler's identity from maclauren and the sine/cosine derivatives, which are different between the version taking radians and the one taking degrees would be more than enough for a quick intuition for me. As you said, many ways. – gtsiam Jun 25 '20 at 15:22
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The degree is a very arbitrary unit. It is basically just 1/360 of a circle. I believe this unit was developed to be based on the rotation of the Earth around the sun. Past measurement techniques approximated the orbit to be about 360 days.

The radian is a much more specific measurement that tends to be easier for conversion and calculation. One radian is defined as the angle of a circle subtended by the radius and an arc length equal to the radius. So for example, in a circle of radius 1 meter, the angle subtended by an arc of length 2 meters is 2/1 = 2 radians.

The questions mathematicians usually answer correlate with the radian instead of the degree.

CountWolves
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