The phase difference between the following two waves $y_2$ and $y_1$ is:

$y_1=a\sin \left(\omega t-kx\right)$

$y_2=b\cos \left(\omega t-kx+\frac{\pi }{3}\right)$

Given: $y_1=a \sin \left(\omega t-kx\right)$ and $y_2=b \cos \left(\omega t-kx+\frac{\pi }{3}\right)$

$y_2=b \cos \left(\omega t-kx+\frac{\pi }{3}\right)=b \sin \left(\omega t-kx+\frac{\pi }{3}+\frac{\pi }{2}\right)$

$y_2=b \sin \left(\omega t-kx+\frac{5\pi }{6}\right)$

The phase difference between $y_2=b sin \left(\omega t-kx+\frac{5\pi }{6}\right)$ and $y_1=a \sin \left(\omega t-kx\right)$ is equal to $\frac{5\pi }{6}$.