Voltage Summing Circuit

From the diagram above, applying Ohm's law gives:

$V_1 - V_P = I_1R_1$ $V_2 - V_P = I_2R_2$ $V_P-V_{\mathrm{out}} = IR_3$

Applying Kirchoff's current law and the current condition for an ideal Op-Amp gives:

$I = I_1 + I_2$

It is assumed that $V_G = 0$ . Using the ideal Op-Amp voltage condition:

$V_P = V_G = 0$

Using all expressions to solve for $V_{\mathrm{out}}$ gives:

$V_{\mathrm{out}} = -\left(\frac{V_1}{R_1}+\frac{V_2}{R_2}\right)R_3$

Given that:

$V_1 = -\cos\left(\omega t\right)$ $V_2 = \sin\left(\omega t\right)$

Leads to:

$V_{\mathrm{out}}= 3\left(\cos\left(\omega t\right)-\frac{\sin\left(\omega t\right)}{2}\right)$ $\implies V_{\mathrm{out}}= \frac{3}{2}\left(2\cos\left(\omega t\right)-\sin\left(\omega t\right)\right)$ $\implies V_{\mathrm{out}}= \frac{3\sqrt{5}}{2}\left(\frac{2}{\sqrt{5}}\cos\left(\omega t\right)-\frac{1}{\sqrt{5}}\sin\left(\omega t\right)\right)$ $\implies V_{\mathrm{out}}= \frac{3\sqrt{5}}{2}\cos\left(\omega t + \arccos\left(\frac{2}{\sqrt{5}}\right)\right)$

Which means that:

$\boxed{M = \frac{3\sqrt{5}}{2} \ ; \ \theta = \arccos\left(\frac{2}{\sqrt{5}}\right)}$

The answer is:

$\boxed{M + \theta \approx 3.81775}$

Where $\theta$ is entered in radians. It would be useful to mention the units of angle to be entered in the problem statement.