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(B) The standard expression for an amplitude modulated $(AM)$ wave is given by: $v_{AM} = (v_0 + A_m \sin \omega_m t) \sin \omega_c t$ Given the messa

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$v_0 \sin \omega_0 t + \frac{A}{2} \sin (\omega_0 - \omega)t + \frac{A}{2} \sin (\omega_0 + \omega)t$

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(B) The standard expression for an amplitude modulated $(AM)$ wave is given by: $v_{AM} = (v_0 + A_m \sin \omega_m t) \sin \omega_c t$ Given the message signal is $A \cos \omega t$ and the carrier wave is $v_0 \sin \omega_0 t$,the modulated signal is: $v_{AM} = (v_0 + A \cos \omega t) \sin \omega_0 t$ $v_{AM} = v_0 \sin \omega_0 t + A \cos \omega t \sin \omega_0 t$ Using the trigonometric identity $2 \sin A \cos B = \sin(A + B) + \sin(A - B)$,we can write: $A \cos \omega t \sin \omega_0 t = \frac{A}{2} [2 \sin \omega_0 t \cos \omega t] = \frac{A}{2} [\sin(\omega_0 + \omega)t + \sin(\omega_0 - \omega)t]$ Substituting this back into the equation: $v_{AM} = v_0 \sin \omega_0 t + \frac{A}{2} \sin(\omega_0 + \omega)t + \frac{A}{2} \sin(\omega_0 - \omega)t$ This matches option $B$.

$A$ signal $A \cos \omega t$ is transmitted using $v_0 \sin \omega_0 t$ as a carrier wave. The correct amplitude modulated $(AM)$ signal is:

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