$12.1.4$ Two sinusoidal waves with the same polarization $E_1~\sin{[\omega (t-z/c)+\varphi_1]}$, $E_2~\sin{[\omega (t-z/c)+\varphi_2]}$ are superimposed on each other. What is the amplitude of the electric field strength of the resulting wave? What is the phase of this wave?
Since waves are superimposed, $$E_R = E_{w1} + E_{w2}$$ $$E_R = E_1~\sin{[\omega (t-z/c)+\varphi_1]} + E_2~\sin{[\omega (t-z/c)+\varphi_2]}$$ $$E_R = (E_1\cos{\varphi_1}+E_2\cos{\varphi_2})\sin{[\omega (t-z/c)]} + (E_1\sin{\varphi_1}+E_2\sin{\varphi_2})\cos{[\omega (t-z/c)]}$$ Let's suppose that $$E_1\cos{\varphi_1}+E_2\cos{\varphi_2} = E \cos{\varphi} = E_x$$ and $$E_1\sin{\varphi_1}+E_2\sin{\varphi_2} = E \sin{\varphi} = E_y$$ and considering the trigonometric identity $\sin{(x+y)} = \sin{x}\cos{y} + \cos{x}\sin{y}$, $$E_R = E \sin{[\omega (t-z/c)+\varphi]}$$ As $E = \sqrt{{E_x}^2 + {E_y}^2}$ and taking in account that $\cos{(x-y)} = \cos{x}\cos{y} + \sin{x}\sin{y}$
$$E = \sqrt{E_1^2+E_2^2+2E_1E_2\cos{(\varphi_1-\varphi_2)}}$$
Phase difference is $$\tan{\varphi} = \frac{E_y}{E_x} = \frac{E_1\sin{\varphi_1}+E_2\sin{\varphi_2}}{E_1\cos{\varphi_1}+E_2\cos{\varphi_2}}$$ Finally, the phase is,
$$\Phi = \omega (t-z/c) + \arctan {\frac{E_1\sin{\varphi_1}+E_2\sin{\varphi_2}}{E_1\cos{\varphi_1}+E_2\cos{\varphi_2}}}$$