Two waves having same amplitude A , angular frequency ω and...

Let the equations of the two waves traveling in opposite directions be $y_1 = A\sin(\omega t - kx)$ and $y_2 = A\sin(\omega t + kx + \phi)$.

Given that the phase difference is \phi = 60^{}^{\circ}.

The resultant displacement is given by the principle of superposition:

$y = y_1 + y_2$

y = A\sin(\omega t - kx) + A\sin(\omega t + kx + 60^{}^{\circ})

Using the trigonometric identity $\sin C + \sin D = 2\sin \left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right), we get:<br><br>$ y = 2A\\sin \left(\dfrac{\omega t - kx + \omega t + kx + 60^{}^{\circ}}{2}\right)\cos \left(\dfrac{\omega t + kx + 60^{}^{\circ} - (\omega t - kx)}{2}\right)<br><br>y = 2A\sin(\omegat + 30^{}^{\circ})\cos(kx + 30^{}^{\circ})$

Answer: 2A\sin(\omegat + 30^{}^{\circ})\cos(kx + 30^{}^{\circ})