If two waves y_1 = A (ω t - 0.1x) and y_2 = A ( - 0.1x - (φ)/(2))...

The given waves are $y_1 = A\sin(\omega t - 0.1x)$ and y_2 = A\sin \left(\omega$t - 0.1x -$\dfrac{\phi}{2}\right).

The phase difference between$the$two waves is \Delta\phi = \dfrac{\phi}{2}.

The resultant amplitude$A_R$ of two superimposed waves is given by:

$A_R =$\sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos(\Delta\phi)}

Substituting $A_1 = A$, $A_2 = A, and$\Delta\phi = \dfrac{\phi}{2}:

$A_R =$\sqrt{A^2 + A^2 + 2A^2\cos\left(\dfrac{\phi}{2}\right)}

$A_R =$\sqrt{2A^2\left(1 + \cos\dfrac{\phi}{2}\right)}

Using$the trigonometric identity1 + \cos$\theta =$2\cos^2 \left(\dfrac{\theta}{2}\right)$:

$A_R = \sqrt{2A^2\left(2\cos^2\dfrac{\phi}{4}\right)}$

$A_R = \sqrt{4A^2\cos^2\dfrac{\phi}{4}}$

$A_R = 2A\cos\dfrac{\phi}{4}$

Answer: $2A\cos\dfrac{\phi}{4}$