Waves on waves

The following statement is generally true (I know this from having studied 3-phase electric power):

$\large{1\angle \theta - 1\angle (\theta - 120^\circ) = \sqrt{3} \angle (\theta + 30^\circ)}$

In this case:

$\large{\theta + 30^\circ = 45^\circ \implies \theta = 15^\circ = \boxed{\frac{\pi}{12} rad}}$

Note: This easy approach assumes a unique solution but doesn't bother to prove it.

Let the two waves be described by $y_1 = acos(ωt + φ_1)$ , $y_2 = acos(ωt + φ_2)$ then the new waves by mean of the identity $\cos(a) + \cos(b)$ = 2 $\cos(\frac{a-b}{2})\cos(\frac{a+b}{2})$ we get that the formula of the new wave is 2a $\cos\frac{φ_1-φ_2}{2}$ $cos(ωt+\frac{φ_1+φ_2}{2})$ . Thus: $\sqrt3 a = 2a\cos\frac{φ_1-φ_2}{2}$ $\Rightarrow$ $φ_1-φ_2 = \pi/3$ and $cos(ωt+\frac{φ_1+φ_2}{2})$ = $\cos(ωt + \pi/4)$ $\Rightarrow$ $φ_1+φ_2 = \pi/2$ Solving the system we get that $φ_2= \pi/12$ and then one of the two initial waves is $y = acos(ωt + \pi/12)$