Equations of motion in the same direction are given by aligned &...

The given equations of motion are $$\begin{aligned} & y_1=2 a \sin (\omega t-k x) \\ & y_2=2 a \sin (\omega t-k x-\theta)\end{aligned}$$ Now the resultant equation of wave is given by $$\begin{aligned} & y=y_1+y_2 \\ & =2 a \sin (\omega t-k x)+2 a \sin (\omega t-k x-\theta) \\ & y=2 a\left[2 \sin \frac{(\omega t-k x+\omega t-k x-\theta)}{2}\times \cos \frac{\omega t-k x-(\omega t-k x-\theta)}{2}\right]\end{aligned}$$y=4 a \cos \frac{\theta}{2} \sin $\left(\omega t-k x-\frac{\theta}{2}\right)$ ...(1) Now comparing equation (1) with $y=A \sin (\omega t-k x)$, we have Resultant amplitude $A=4 a \cos \frac{\theta}{2}$