How can we find the signal power of \$x_{out}(t)=\frac{1}{2}Am(t)cos \theta\$?
$$x_{out}(t)=\frac{1}{2}Am(t)cos \theta$$
I can group constants together and write this as:
$$x_{out}(t)=(\frac{1}{2}Acos \theta V_m)cos\omega_m t$$
By the general definition of power of a sinusoidal wave,
$$P_1 = \frac{(\frac{1}{2}Acos \theta V_m)^2}{2}$$
$$=\frac{1}{8}A^2V_m^2cos^2\theta$$
Now \$x(t)=Am(t)cos (\omega_c t)\$,
$$\implies x(t) = AV_mcos(\omega_mt)cos(\omega_ct) ----(1)$$

I can write x(t) as sum of two sinusoids:
$$\implies x(t) = \frac{AV_m}{2}(cos(\omega_c+\omega_m)t+cos(\omega_c-\omega_m)t)$$
$$\implies x(t) = \frac{AV_m}{2}cos(\omega_c+\omega_m)t+\frac{AV_m}{2}cos(\omega_c-\omega_m)t$$

These represents two side bands of the DSB-SC modulated wave. Sum of the sideband powers will be the total power of the modulated wave.

$$\implies P_2 = \frac{A^2V_m^2}{8} + \frac{A^2V_m^2}{8} = \frac{1}{4}A^2V_m^2 $$

$$\therefore P_1/P_2 = \frac{1}{2} cos^2\theta$$