For the following question: $$\cos(\omega t)+2\cos\left(\omega t+\frac{\pi}{4}\right)+3\cos\left(\omega t+\frac{\pi}{2}\right)$$

\begin{align} & 1\angle0^\circ + 2\angle45^\circ+3\angle90^\circ\\ =& 1+\sqrt2+j\sqrt2+3j\\ =& 1+\sqrt2 + j\left(3+\sqrt2\right) \end{align}

Which has

$$A\approx5.03, \phi\approx1.07\text{rad}$$

Thus answer is $5\angle61^\circ$, or $5\cos(\omega t+1.07)$.

This is the correct answer, but what confuses me is that to use phasor form, don't we need both sine and cosine according to eulers identity? Why were we able to convert a cosine without a sine to a phasor and then convert it back to just a cosine without a sine? Eulers identity says that $Ae^{j\theta} = A\cos\theta - jA\sin\theta$

Thank you for anything you can provide :)

• I think I'm horribly missing a concept here. I tried reading through the section in the textbook but still don't get it. – user96193 Sep 22 '13 at 15:16

Hint: Compare your original question, the real part of $($phasor$\times e^{j\omega t})$, and your answer.
\begin{align} e^{j\omega t} + 2e^{j\omega t+j\frac{\pi}{4}} + 3 e^{j\omega t+j\frac{\pi}{2}} =& e^{j\omega t}\left(1+2e^{j\frac{\pi}{4}}+3e^{j\frac{\pi}{2}}\right)\\ =& e^{j\omega t}\left(1+\sqrt2 + j\sqrt2+j3\right)\\ \approx& e^{j\omega t}\cdot5.03e^{j1.07}\\ =& 5.03 e^{j\omega t+j1.07}\\ \Re\left(e^{j\omega t} + 2e^{j\omega t+j\frac{\pi}{4}} + 3 e^{j\omega t+j\frac{\pi}{2}}\right) \approx& \Re\left(5.03 e^{j\omega t+j1.07}\right)\\ =& \Re\left[5.03 \cos(\omega t+1.07)+j5.03\sin(\omega t+1.07)\right]\\ =& 5.03 \cos(\omega t+1.07) \end{align}