Proof of Euler's Formula with No Angle Addition

[Deveno]

As I see it, you've proved:

$e^{iz} = \cos\theta + i \sin\theta$ for some $\theta \in [0,2\pi)$

I see no problem with assuming $z \in \Bbb R$ but I think you still have to show:

$z - \theta = 2k\pi,\ k \in \Bbb Z$

 

[mathbalarka]

Ah, you're right, we are at the point of differentiating sin and cos again, true. :p

 

[Deveno]

Recall that differentiating geometrically is the limit of a certain ratio (of two differences), and that trig functions are themselves ratios. Can you show geometrically that:

$(\sin)'(x) = \cos x$?

(I believe you CAN if you have the angle-sum formulas already established...but...)

 

[mathbalarka]

Yes, I had a proof for it in my mind and was trying to prove it the other way around. Nevermind, here it is :

The beginning of the proof of mine starts similarly as chisigma's : Let $\vec{r}$ be a point moving along the unit circle with uniform velocity and with co-ordinate $r(t) = (\cos(t), \sin(t))$. We have $|\vec{r}| = 1$ as well as $\vec{r} \cdot \vec{r} = 1$.

Differentiating, we obtain

$$\vec{\frac{dr}{dt}} \cdot \vec{r} + \vec{r} \vec{\cdot \frac{dr}{dt}} = 0$$

i.e., $\vec{r} \cdot \vec{\frac{dr}{dt}} = 0$. This implies the vectors $\vec{r}$ and $\vec{\frac{dr}{dt}}$ are orthogonal. Then all the co-ordinates of $\vec{r}$ are shifted by a sum of $\pi/2$. Hence,

$$ r'(t) = (\cos(t + \pi/2), \sin(t + \pi/2)) = (-\sin(t), \cos(t))$$

Note that these are obtained by noting the symmetries of circle, so has nothing to do with the angle-sum formula. Note further that the equality still doesn't follow as we have no knowledge about the scalar $\left | \vec{\frac{dr}{dt}} \right |$. This can be proved to be 1 by noting that the it is the velocity of the vector $r$ which is in turn equals to $\frac{d}{T}$ where $d$ is the distance, i.e, $2\pi$ and $T$ is the period of motion, i.e, $2\pi$. This proves that

$$\frac{d}{dt} \left ( \sin(t) \right ) = \cos(t)$$
$$\frac{d}{dt} \left ( \cos(t) \right ) = -\sin(t)$$

Does this look good?

 

[Deveno]

Yes, I had a proof for it in my mind and was trying to prove it the other way around. Nevermind, here it is :

The beginning of the proof of mine starts similarly as chisigma's : Let $\vec{r}$ be a point moving along the unit circle with uniform angular speed $\frac{1}{2\pi}$ and with co-ordinate $r(t) = (\cos(t), \sin(t))$. We have $|\vec{r}| = 1$ as well as $\vec{r} \cdot \vec{r} = 1$.

Differentiating, we obtain

$$\vec{\frac{dr}{dt}} \cdot \vec{r} + \vec{r} \vec{\cdot \frac{dr}{dt}} = 0$$

i.e., $\vec{r} \cdot \vec{\frac{dr}{dt}} = 0$. This implies the vectors $\vec{r}$ and $\vec{\frac{dr}{dt}}$ are orthogonal. Then all the co-ordinates of $\vec{r}$ are shifted by a sum of $\pi/2$. Hence,

$$ r'(t) = (\cos(t + \pi/2), \sin(t + \pi/2)) = (-\sin(t), \cos(t))$$

Note that these are obtained by noting the symmetries of circle, so has nothing to do with the angle-sum formula. Note further that the equality still doesn't follow as we have no knowledge about the scalar $\left | \vec{\frac{dr}{dt}} \right |$. This can be proved to be 1 by noting that the it is the velocity of the vector $r$ which is in turn equals to $\frac{d}{T}$ where $d$ is the distance, i.e, $2\pi$ and $T$ is the period of motion, i.e, $2\pi$. This proves that

$$\frac{d}{dt} \left ( \sin(t) \right ) = \cos(t)$$
$$\frac{d}{dt} \left ( \cos(t) \right ) = -\sin(t)$$

Does this look good?

A couple of minor notes:

It is the arguments of the coordinates that are shifted by $\dfrac{\pi}{2}$.

We can't say "which way" the shift occurs from your argument, it could be plus or minus, yes? Also, orthogonality only gives $r'(t)$ up to a scalar, so your FIRST equation involving it is not quite correct (which is why you need to find $|r'(t)|$). So I think you still need to show whether we have:

$r'(t) = (\cos t, -\sin t)$ or:
$r'(t) = (-\cos t, \sin t)$

 

[mathbalarka]

So I think you still need to show whether we have:

$r'(t) = (\cos t, -\sin t)$ or:
$r'(t) = (-\cos t, \sin t)$

As $\vec{\frac{dr}{dt}}$ is the velocity vector of the point, it has the same direction as the position vector $\vec{r}$. Hence, no sign change is possible.