I'm facing much confusion about a rather silly problem recently. Suppose I have two functions $\sin(x+\phi_1)$ and $\sin(x+\phi_2)$.
Defining $\theta_1=x+\phi_1$ as the phase of the first and $\theta_2=x+\phi_2$ as the phase of the second function.
The phase difference, by definition comes out to be simply:
$$\Delta\theta=\theta_1-\theta_2=x+\phi_1-x-\phi_2=\phi_1-\phi_2$$
Hence, $$\Delta\theta=\phi_1-\phi_2$$
So, there is a constant phase difference between these two waves.
However, what if I have two functions $\sin(x+\phi_1)$ and $\sin(\phi_2-x)$ instead ? If I follow the same definition above, then I have $\theta_1=x+\phi_1$ and $\theta_2=\phi_2-x$.
In this case, the phase difference comes out to be $$\Delta \theta=2x+(\phi_1-\phi_2)$$
As $x$ increases, the phase difference seems to increase here. However, this doesn't seem correct to me.
For example, I know $\cos(x)=\sin(\frac{\pi}{2}-x)=\sin(\frac{\pi}{2}+x)$. This automatically implies, that both $\sin(\frac{\pi}{2}-x)$ and $\sin(\frac{\pi}{2}+x)$ have the same phase, since they are both equal to $\cos(x)$.
However if I use the definiton as above, then $\theta_1=\frac{\pi}{2}-x$ and $\theta_2=\frac{\pi}{2}+x$. In this case however,
$$\theta_1-\theta_2=2x=2\omega t \,\,\,\,\,\,(x=\omega t)$$
However, this implies that the phase difference increases over time. When the difference is an even multiple of $\pi$, the waves are in phase, and apart from that, they are out of phase. However, in this example, since both are essentially the same function, we should have a obtained a constant phase difference that was always $0$.
What is wrong with my definition then, and how would I find the phase difference between two waves in general.
Best Answer
In calculating the phase difference between two waves, it seems necessary that both waves have the same functionality as a function of the independent variable and that the frequency of the two waves are the same, i.e., their period is the same. In that case the phase difference may be interpreted as the horizontal shift that is necessary to make the two waves fall right on top of each other (assuming they have the same amplitude). It is this fact that necessitates the same period of the two waves. Once the periods are the same, the shift necessary to make the two waves identical is definitely less than the equal wave period. With that, one can proceed to each of your examples to calculate the phase difference. In the first case the two waves are $\sin(x+\phi_{1})$ and $\sin(x+\phi_{2})$. Note that the two waves have the same exact frequency and both are the same periodic function. Thus the magnitude of the phase difference is $$|(x+\phi_{1})-(x+\phi_{2})|=|\phi_{1}-\phi_{2}|$$
In the second case the two waves are $\sin(x+\phi_{1})$ and $\sin(\phi_{2}-x)$. While both waves are $\sin$ functions, they do not have the exact same frequency (though the absolute value is the same). Thus, as mentioned in the comments $$\sin(\phi_{2}-x)=-\sin(x-\phi_{2})=\sin(x-\phi_{2}+\pi)$$ Subsequently, the magnitude of the phase difference in this case is $$|(x+\phi_{1})-(x-\phi_{2}+\pi)|=|\phi_{1}+\phi_{2}-\pi|$$ Note that as mentioned in your problem, the implication of the above equation for $\cos(x)=\sin(\pi/2-x)=\sin(\pi/2+x)$ is that the phase difference between $\sin(\pi/2-x)$ and $\sin(\pi/2+x)$ is 0. I think this answer provides general guidelines for finding the phase difference between two waves through specific examples mentioned in the question. Please let me know if you have any questions or find the answer incorrect.