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+### Условие
+
+$12.1.4.$ [Вставьте условие задачи]
+
+### Решение
+
+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 $\cos{(x-y)} = \cos{x}\cos{y} + \sin{x}\sin{y}$
+
+#### Answer 1
+
+$$
+E = \sqrt{E_1^2+E_2^2+2E_1E_2\cos{(\varphi_1-\varphi_2)}}
+$$
+
+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,
+
+#### Answer 2
+
+$$
+\Phi = \omega (t-z/c) + \arctan {\frac{E_1\sin{\varphi_1}+E_2\sin{\varphi_2}}{E_1\cos{\varphi_1}+E_2\cos{\varphi_2}}}
+$$
+
+#### Ответ
+
+[Вставьте краткий ответ или результат в рамке]