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Let us start with the obvious.

I know the formulae for angles. I know how to apply them. I also know the formulae involving $e$.

But I don't understand what sine has to do with Euler's $e$. (Neither do I for cosine or tangent)

If you were to build a course that relies on truly understanding those three functions and to a certain degree their implications, where would you start?

Hans Lundmark
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SAJW
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2 Answers2

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Well, we need a definition for $e^x$, which is the only part your missing. And...I choose this one!

$$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n$$

And it just so happens that if I let $x\to ix$, I get the following:

$$e^{ix}=\lim_{n\to\infty}\left(1+\frac{ix}n\right)^n$$

An animation of this for $x\in[0,\pi)$

enter image description here

Interestingly, it approaches a circle, which, if we remember the Cartesian coordinates on the unit circle:

$$e^{ix}=\cos(x)+i\sin(x)$$

Simply Beautiful Art
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  • Note that the animation, specifically, is for $x=\pi$. – Akiva Weinberger Jan 23 '17 at 03:28
  • @AkivaWeinberger well, it was borrowed from a Wikipedia purely about $e^{i\pi}+1=0$. – Simply Beautiful Art Jan 23 '17 at 12:23
  • In order to illustrate a trajectory as a function of $t,$ you have to start plotting the trajectory somewhere and stop somewhere. Starting at $t=0$ and stopping when you reach the estimated position at $t=\pi$ is a happy choice even if your main purpose is not to prove Euler's identity. The animation illustrates how the change in radius along the trajectory goes to zero as $n\to\infty.$ – David K Feb 25 '17 at 16:04
  • You are implicitly using the fact that the limit definition of $e^{ix}$ leads to your last equation. The proof although not difficult is non-trivial. You can have a look at http://math.stackexchange.com/a/2161021/72031 or http://math.stackexchange.com/a/1668179/72031 – Paramanand Singh Feb 25 '17 at 16:41
  • @ParamanandSingh Thanks. When looking at this problem, I suppose I trade triviality for intuition. – Simply Beautiful Art Feb 25 '17 at 16:43
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It's a bit of a vague question. Euler's is usually the relation we think of when relating trig functions to $e$. It states:

$$e^{i\theta} = \cos(\theta) + i\sin(\theta)$$

The typical proof involves Taylor Series. If you don't know about Taylor Series an "easier" way is to prove it:

  1. Verify the initial conditions are the same.
  2. Show that both $e^{i\theta}$ and $\cos(\theta) + i\sin(\theta)$ satisfy:

$$D_zf(z) = if(z)$$

The full proof can be found here. But you should try working it out yourself.

A more geometric interpretation (however, not really a proof) is that both $e^{i\theta}$ and $\cos(\theta) + i\sin(\theta)$ both represent the circle on the complex plane.

Dair
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  • Not sure how they'd both represent a circle on the complex plane if you haven't defined $e^{i\theta}$ yet. – Simply Beautiful Art Jan 22 '17 at 23:08
  • @SimplyBeautifulArt It's not part of the proof. Just intuition. The formal proof is in the first part. – Dair Jan 22 '17 at 23:09
  • @SimplyBeautifulArt To put another way, you prove Euler's formula, and then you notice that $\cos(\theta) + i\sin(\theta)$ is a circle, and thus $e^{i\theta}$ describes a circle in the complex plane. – Dair Jan 22 '17 at 23:12
  • I don't want to speak down to this post. it was probably in the best intentions. But for me truly understanding something is being able to imagine it. But maybe i make up my mind in the future and see the value in your answer. Don't be deterred by my ignorance ;). I want to add that I upvoted, just not accepted this as an answer. – SAJW Jan 22 '17 at 23:17