Proof of $\sin^2 x+\cos^2 x=1$ using Euler's Formula How would you prove $\sin^2x + \cos^2x = 1$ using Euler's formula?
$$e^{ix} = \cos(x) + i\sin(x)$$
This is what I have so far:
$$\sin(x) = \frac{1}{2i}(e^{ix}-e^{-ix})$$
$$\cos(x) = \frac{1}{2} (e^{ix}+e^{-ix})$$

$e^{ix}=\cos x+i\sin x$
Putting $x=y$ and $x=-y$ respectively, $ e^{iy}=\cos y+i\sin y$ and $e^{-iy}=\cos(-y)+i\sin(-y)=\cos y-i\sin y$
So, $(\cos y+i\sin y)(\cos y-i\sin y)=e^{iy}e^{-iy}$
$\implies \cos^2y+\sin^2y=1$
or, $$(\cos y)^2+(\sin y)^2$$
$$=\left(\frac{e^{iy}+e^{iy}}{2}\right)^2+\left(\frac{e^{iy}-e^{-iy}}{2i}\right)^2$$
$$=\frac{(e^{iy}+e^{iy})^2-(e^{iy}-e^{iy})^2}4=\frac{4e^{iy}e^{-iy}}4=\frac44$$
