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## Homework Statement

Using the imaginary parts

When using complex representation, it is customary to use the real parts. Instead use the imaginary part of [itex]e^{j\theta}[/itex] to calculate an expression for the sum:

[tex]\sin(\omega t) + \sin((\omega + \Delta \omega)t)[/tex]

Remember, it should come out to be the beats formula.

## Homework Equations

[tex]

e^{j\theta} = \cos(\theta) + j\sin(\theta)\\

\omega_{beat} = |\omega_{1}-\omega_{2}|\\

cos(\theta) = Re[e^{j\theta}]\\

sin(\theta) = Im[e^{j\theta}]

[/tex]

## The Attempt at a Solution

I'm having trouble understanding what the question is asking for. I tried rewriting the above equation using the complex exponential with the hope to take the imaginary part later but that hasn't seemed to bring me closer to the "beat formula". I'm also not sure what it means "it should come out to be the beats formula"

[tex]

\sin(\omega t) + \sin((\omega+\Delta \omega)t)\\

\sin(\omega_{1}) + \sin(\omega_{2}t)\\

e^{j\omega_{1}t} + e^{j\omega_{2}t}\\

[/tex]

Using some trig I get an expression for the sum that does have look like it will exhibit beating behavior when the frequencies are close, but this wasn't using the complex exponential function.

[tex]

\sin(\alpha) + \sin(\beta) = 2 \sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)\\

\sin(\omega_{1}t)+\sin(\omega_{2}t)= 2\sin\left(\frac{\omega_{1}+\omega{2}}{2}t\right)\cos\left(\frac{\omega_{1}-\omega_{2}}{2}t\right)

[/tex]

The biggest question I'd like to ask is: How do you use the imaginary part of [itex]e^{j\theta}[/itex] to calculate an expression for the sum?