Recent content by ke7ijo

Thank you! That's it.

I differentiated both sides of Euler's formula with respect to x : e^ix = sin x + i cos x => ie^ix = cos x - i sin x Then for comparison I multiplied both sides of Euler's formula by i: e^ix = sin x + i cos x => ie^ix = i sin x - cos x Each of these two procedures seems to yield the...