Why is $i\cdot \sin(x)$ not $\cos(x)$?

Question

I recently repeated some math basics of the Fourier transform and of course stumbled across Euler's formula. When reading the term $\cos(x) + i\sin(x)$ I wondered why it could not be written as $2\cos(x)$. Since all professors always emphasize that a cosine is nothing but a $90$ degree shifted sine, I was wondering why the multiplication with i, which also causes a $90°$ shift on the complex plane, doesn't result in a $\cos$-function.

Answer 1

"Cosine is 90 degree shifted sine" means a shift in the arguments, not the rotation of the values on the complex plane. That is $ \cos(x) = \sin(x+90^\circ) $. Multiplication by $i$ is rotating the value of the function, not shifting the argument.

Answer 2

It causes a shift when we deal with angles and not with functions of angles.

For example, in a frequency domain, we may write, $5i$ as $5\angle90^o$, which is a notation for $5e^{(i\omega t +\pi/2)}$

But here, we have $i. sin(x)$ and not a magnitude of a time or frequency domain voltage or current or whatsoever.

So, $i\cdot\sin(x)$ is not equal to $cos(x)$ . Also only $\sin(90°\pm x) = \cos(x)$, not when we have $i\cdot \sin(x)$.