# Phasor/Harmonic Addition Formula/Theorem: Why can we take out the frequency out of an complex argument?

Those four name can be used as a keyword on google.
I haven't known the official name and think that they seems to be equivalent each other.

From this paper on page 3

A Hyperbolic Analog of the Phasor Addition Formula by F. Adrián F. Tojo (July 30, 2018)

Figure 1. Part of page 3 of the paper mentioned.

If I rewrite that,

$$\displaystyle \mathrm{a} \ \mathrm{e}^{j \alpha} + \mathrm{b} \ \mathrm{e}^{j \beta} = \sqrt{ \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left( \alpha - \beta \right) } \quad \mathrm{e}^{\ j \ \arg\left[ \left[ \mathrm{a} \cos\left( \alpha \right) + \mathrm{b} \cos\left( \beta \right) \right] + j \left[ \mathrm{a} \sin\left( \alpha \right) + \mathrm{b} \sin\left( \beta \right) \right] \right]}$$

So if we take the real part of it, we get

$$\small \displaystyle \mathrm{a}\cos\left( \alpha \right) + \mathrm{b} \cos\left( \beta \right) = \sqrt{ \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left( \alpha - \beta \right) } \: \cos\left[ \mathrm{arg}\left[ \left[ \mathrm{a} \cos\left( \alpha \right) + \mathrm{b} \cos\left( \beta \right) \right] + j \left[ \mathrm{a} \sin\left( \alpha \right) + \mathrm{b} \sin\left( \beta \right) \right] \right] \right]$$

Suppose if $$\alpha = \omega t + \phi_1$$ and $$\beta = \omega t + \phi_2$$, then

\begin{align} \small \displaystyle \mathrm{a}\cos\left( \omega t + \phi_1 \right) + \mathrm{b} \cos\left( \omega t + \phi_2 \right) &= \sqrt{ \small \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left(\phi_1 - \phi_2 \right) } \\ &\cdot\cos\left[ \small \mathrm{arg}\left[ \left[ \mathrm{a} \cos\left( \omega t + \phi_1 \right) + \mathrm{b} \cos\left( \omega t + \phi_2 \right) \right] + j \left[ \mathrm{a} \sin\left( \omega t + \phi_1 \right) + \mathrm{b} \sin\left( \omega t + \phi_2 \right) \right] \right] \right] \end{align}

Why can we take the frequency out of the complex argument like this?

\begin{align} \small \displaystyle \mathrm{a}\cos\left( \omega t + \phi_1 \right) + \mathrm{b} \cos\left( \omega t + \phi_2 \right) &= \sqrt{ \small \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left(\phi_1 - \phi_2 \right) } \\ &\cdot\cos\left[ \small \omega t + \mathrm{arg}\left[ \left[ \mathrm{a} \cos\left( \phi_1 \right) + \mathrm{b} \cos\left( \phi_2 \right) \right] + j \left[ \mathrm{a} \sin\left( \phi_1 \right) + \mathrm{b} \sin\left( \phi_2 \right) \right] \right] \right] \end{align}

$$\displaystyle \mathrm{a} \, \mathrm{e}^{j \alpha} + \mathrm{b} \, \mathrm{e}^{j \beta} = \left[ \mathrm{a} \cos\left( \alpha \right) + \mathrm{b} \cos\left( \beta \right) \right] + j \left[ \mathrm{a} \sin\left( \alpha \right) + \mathrm{b} \sin\left( \beta \right) \right] \label{1}\tag{1}$$

Remember that the complex argument form inside of the cosine is equivalent to \eqref{1}.
Or just use Euler's formula, it's the same.

$$\small \displaystyle \mathrm{a} \, \cos\left( \omega t + \phi_1 \right) + \mathrm{b} \, \cos\left( \omega t + \phi_2 \right) = \sqrt{ \small \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left(\phi_1 - \phi_2 \right) } \cdot\cos\left[ \small \mathrm{arg}\!\left[ \mathrm{a} \, \mathrm{e}^{j \left( \omega t + \phi_1 \right)} + \mathrm{b} \, \mathrm{e}^{j \left( \omega t + \phi_2 \right)} \right] \right]$$

Factor out the frequency

$$\small \displaystyle \mathrm{a} \, \cos\left( \omega t + \phi_1 \right) + \mathrm{b} \, \cos\left( \omega t + \phi_2 \right) = \sqrt{ \small \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left(\phi_1 - \phi_2 \right) } \cdot\cos\left[ \small \mathrm{arg}\!\left[ \mathrm{e}^{j \omega t} \left( \mathrm{a} \, \mathrm{e}^{j \phi_1 } + \mathrm{b} \, \mathrm{e}^{j \phi_2 } \right) \right] \right]$$

Remember the complex argument identities

$$\mathrm{arg}\!\left(z_1 z_2\right) = \mathrm{arg}\!\left(z_1\right) + \mathrm{arg}\!\left(z_2\right)$$

And also the fact that

$$\mathrm{arg}\!\left(\mathrm{e}^{j \theta}\right) = \theta$$

Thus

\begin{align} \small \displaystyle \mathrm{a} \, \cos\left( \omega t + \phi_1 \right) + \mathrm{b} \, \cos\left( \omega t + \phi_2 \right) &= \sqrt{ \small \mathrm{a}^2 + \mathrm{b}^2 + 2 \ \mathrm{a} \ \mathrm{b} \ \cos\left(\phi_1 - \phi_2 \right)} \\ &\cdot\cos\left[ \small \omega t + \mathrm{arg}\!\left[ \left[ \mathrm{a} \cos\left( \phi_1 \right) + \mathrm{b} \cos\left( \phi_2 \right) \right] + j \left[ \mathrm{a} \sin\left( \phi_1 \right) + \mathrm{b} \sin\left( + \phi_2 \right) \right] \right] \right] \end{align}

The point is that, if the cosines on the left side has the same phase part which is separated by addition/subtraction sign, we can take out of it from the complex argument function, hence simplifies it.

• Please critic my answer if it's incorrect. May 13, 2019 at 1:16