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(B) The first wave is given by $y_1 = A \sin (\omega t + kx + 0.57)$.The second wave is $y_2 = A \cos (\omega t + kx)$.Using the trigonometric identit

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$1.0 \ 	ext{radian}$

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(B) The first wave is given by $y_1 = A \sin (\omega t + kx + 0.57)$.The second wave is $y_2 = A \cos (\omega t + kx)$.Using the trigonometric identity $\cos 	heta = \sin (	heta + \frac{\pi}{2})$,we can rewrite $y_2$ as:$y_2 = A \sin (\omega t + kx + \frac{\pi}{2})$.The phase of the first wave is $\phi_1 = \omega t + kx + 0.57$.The phase of the second wave is $\phi_2 = \omega t + kx + \frac{\pi}{2}$.The phase difference $\Delta \phi$ is given by $|\phi_2 - \phi_1| = |(\omega t + kx + \frac{\pi}{2}) - (\omega t + kx + 0.57)|$.$\Delta \phi = \frac{\pi}{2} - 0.57$.Since $\frac{\pi}{2} \approx 1.57$,we have:$\Delta \phi = 1.57 - 0.57 = 1.0 \ 	ext{rad}$.

Two waves are represented by the equations $y_1 = A \sin (\omega t + kx + 0.57) \ m$ and $y_2 = A \cos (\omega t + kx) \ m$,where $x$ is in metre and $t$ is in second. What is the phase difference between them?

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