# How does the 'phase' of a cosine wave become 'pi' when it transitions from positive to negative?

I was watching this video lecture on filters where this professor says that:

The phase of a cosine wave $\cos(\omega)$ is 0 till $\omega=\pi/2$. As soon as $\omega$ goes beyond $\pi/2$, i.e when $\cos(\omega)$ becomes negative, its phase becomes $\pi$.

I couldn't understand this. How can $\cos(\omega)$ have phase in itself if it is just a plain normal cosine function? What is the physical interpretation of this phase?

• See my comment below – Jazzmaniac Sep 27 '17 at 18:46

The statement in the video is, at best, misleading. Normally, when people speak of the "phase" of a sinusoidal function $\cos \omega$, they are talking about the value of $\omega$. For instance, when we say that a wave undergoes a "phase shift", we mean that the value of $\omega$ has changed. More information on this definition of the word "phase" is at https://en.wikipedia.org/wiki/Phase_(waves).
As noted by Jazzmaniac in comments, we could also try to interpret the statement by regarding the value of $\cos \omega$ as a complex number $z$. In polar representation we can write $z = |z| e^{i \phi}$ where $\phi$ is the phase of the number $z$. This use of the word "phase" is misleading in this context. For instance, applying a "phase shift" would mean that we multiply $\cos \omega$ by some complex number $e^{i \phi'}$. That certainly wouldn't implement a phase shift of the sinusoidal wave, though, in the usual sense.
• @Schimay, are you familiar with the polar representation of complex numbers? The phase of a complex number is just another name for the argument $\phi$ of its complex representation $z=|z| \exp(i \phi)$. – Jazzmaniac Sep 27 '17 at 19:21
• @AlexTP, it doesn't. Read the original post again. As the argument of the cosine crosses $\pi/2$, the phase jumps from $0$ to $\pi$. – Jazzmaniac Sep 27 '17 at 20:01
• @Schimay To put it another way: a negative number $M$ can be written as $|M|e^{j\pi}$; i.e. every real negative number has phase $\pi$. – MBaz Sep 27 '17 at 20:38