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

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.

• note that for any chosen $x\in\Bbb R$ we have that $i\sin(x)$ is a purely imaginary number, however $2\cos (x)$ is real, so they cannot be equal – Masacroso May 1 at 10:49

"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.

• I encountered another problem: If it is like you say, why is it that differentiation in the time domain (which would make a sine a cosine) is multiplication with i in the frequency domain? This is why the impedances of capacitors and inductors have the i in their formula, isnt it? But if cos(x) is not i*sin(x), how does this make sense? – Philipp317 Jul 23 at 6:38
• @Philipp317 I would need more contest to be sure, but I think you're talking about a case of alternating current, where for example we assume that the charge on the capacitor is $Q(t) = {\rm Re}(Q_0 e^{i\omega t})$. Then you have $$I = \frac{dQ}{dt} = {\rm Re}(i\omega Q_0 e^{i\omega t}) = {\rm Re}(I_0 e^{i\omega t})$$ where $I_0 = i\omega Q_0$. – Adam Latosiński Jul 23 at 9:06
• I was talking about the current-voltage relation for a capacitor. It's described by I = C * dU/dt , basically meaning that if the applied voltage is a sine-wave, the resulting current is a cosine-wave. It effectively means a 90° degree shift for a harmonic input. The expression to describe the impedance of a capacitor in the frequency domain is 1/jwC , and here I thought the division by j in the denominator expresses the same phase shift the differential equation in the time domain described. But the answers told me, those 2 "phase shifts" are not the same – Philipp317 Jul 24 at 15:39
• Division by $i$ is a phase shift in the oposite direction because while $i=e^{i\pi/2}$, we have $\frac{1}{i} = -i = e^{-i\pi/2 }$. – Adam Latosiński Jul 24 at 17:59
• No, you must compare real part to the real part and imaginary part to the imaginary part. That gives you $$-sin(\omega t) = \cos(\omega t + \pi/2)$$ $$\cos(\omega t) = \sin(\omega t+\pi/2)$$ – Adam Latosiński Jul 28 at 7:33

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 $$i\cdot \sin(x)$$.