Cos derivation from the euler's formula
$\cos x = \dfrac{e^{ix} + e^{-ix}}{2}$ usually starts with $e^{-ix} = \cos x - i \sin x$ but it is rarely explained why this is true. Or sometimes it is shown that
$e^{i(-x)} = \cos(-x) + i \sin(-x) = \cos x - i \sin x$
and thats fine but I've never seen a more elegant way which I just came up with.
$e^{-ix} = \dfrac{1}{e^{ix}}$, right? So we move the euler's identity to the denominator
$e^{-ix} = \dfrac{1}{\cos x + i \sin x}$
Now we multiply both sides of the fraction by $\cos x - i \sin x$ to make use of the difference of squares.
We get $e^{-ix} = \dfrac{\cos x - i \sin x }{(\cos x + i \sin x)(\cos x - i \sin x)} = \dfrac{\cos x - i \sin x}{(\cos^2 x - i^2 \sin^2 x)}$
And we know that i^2 = -1 and $\cos^2 x + \sin^2 x = 1$ so finally
$e^{-ix} = \dfrac{\cos x - i \sin x}{1}$
Simple and elegant. Why nobody uses this approach?