How do you get from easy trig functions to complex, exponential and factorial functions

Question

It seems a bit weird to me that $\cos(x)=\frac{e^{ix}+e^{-ix}}2$ which at the same time equals an infinite sum and the hyperbolic cosine with $i$ in it. How can a human think of such complex concepts?

Answer 1

This is easy to show with calculus.

$e^x = \sum_\limits{n=0}^{\infty} \frac {x^n}{n!}\\ \cos x = \sum (-1)^n\frac{x^{2n}}{(2n)!}\\ \sin x = \sum (-1)^n\frac{x^{2n+1}}{(2n+1)!}\\ e^{ix} = \cos x + i\sin x$

But, if you have not learned calculus.

if $z = a+bi$ is a complex number

$|z| = \sqrt {a^2 + b^2}$ is the length of this vector.

For complex $z,w, |zw| = |z||w|$

If we look at only unit vectors.

let $a = \cos \theta, b = \sin\theta \\ x = \cos \phi, y = \sin \phi$

$(a+bi)(x+yi) =$$ (ax - by) + (ay + bx)i\\ \cos(\theta + \phi) + i\sin(\theta + \phi)$

Multiplication of these vectors adds the angle these vectors form with the real axis.

We might also look at: $f(\theta) = cos \theta + i\sin \theta$

$f(\theta+\phi) = (cos \theta\cos\phi + \sin \theta\sin \phi) + i(\sin\theta\cos\phi + \cos\theta\sin\phi) = f(\theta)f(\phi)$

This is a property of an exponential function.

$\exp(\theta+\phi) = \exp(\theta)\exp(\phi)$

The question becomes, what is the base?

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

let $ixn = m$

$e = \lim_\limits {m\to \infty} (1 + \frac {x}{m}i)^m$

as we let $m$ grow we get a picture like:

As $m$ gets larger, this product spirals tighter to the circle. As m gets to be very large, it lies right on the circle, and the length of that arc equals $x.$

Answer 2

$$\exp(xi) = \cos(x) + i \sin(x)$$

\begin{align}\exp(-xi)& = \cos(-x) + i \sin(-x) \\ &=\cos(x)-i\sin(x)\end{align}

Just sum them up and divide it by $2$.

Answer 3

Since$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots,$$you have\begin{align}e^{ix}&=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots\\&=1+ix-\frac{x^2}{2!}-i\frac{x^3}{3!}+\cdots\\&=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots+\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\right)i\\&=\cos(x)+\sin(x)i.\end{align}Therefore, $e^{-ix}=\cos(-x)+\sin(-x)i=\cos(x)-\sin(x)i$. So\begin{align}\frac{e^{ix}+e^{-ix}}2&=\frac{\cos(x)+\sin(x)i+\cos(x)-\sin(x)i}2\\&=\cos(x).\end{align}