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For the equation:

$$A_1 \sin (\omega t + \theta_1) + A_2 \sin (\omega t + \theta_2) = A_3 \sin (\omega t + \theta_3)$$

I've been able to show that the amplitude of the sum is (I believe this is a standard problem):

$$ A_3 = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\theta_1 - \theta_2)} $$

and the new phase is:

$$ \theta_3 = \arctan \left(\frac{A_1 \sin \theta_1 + A_2 \sin \theta_2}{A_1 \cos \theta_1 + A_2 \cos \theta_2}\right) $$

My question is what happens when the phases $\theta_1$ and $\theta_2$ are zero (or just equal to each other). Most books and internet sites (eg http://mathworld.wolfram.com/HarmonicAdditionTheorem.html) state that under these conditions:

$$ A_3 = \sqrt{A_1^2 + A_2^2} $$

However, if I set $\theta_1$ and $\theta_2$ to zero one gets instead (Since $cos(0) = 1$):

$$ A_3 = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2} $$

I've been staring at this for a while and I don't understand why the published answer is different from mine. I am sure it is something simple but I've not been able to spot it (Addendum: Wikipedia site: https://en.wikipedia.org/wiki/List_of_trigonometric_identities has the same result under section Linear Combinations).

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    $\begingroup$ I realized the problem: $A_3 = \sqrt{A^2_1 + A^2_2 + 2 A_1 A_2}$ applies to the relation $cos (wt + \theta_1) + sin(wt + \theta_2)$ not $sin (wt + \theta_1) + sin(wt + \theta_2)$. For the latter it works fine. This is an error on my part. ok to delete this question, as its not a valid one? $\endgroup$ – rhody Jul 13 '18 at 18:23
  • $\begingroup$ Although your Question contained and was motivated by an error, it brings up some useful content, and I'd like to retain it. Even the error, articulated in your Comment, bears pointing out as it is the sort of thing anyone might overlook at first glance. $\endgroup$ – hardmath Jul 14 '18 at 13:02
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Be sure to correctly read your references.

The formula $\sqrt{A_1^2+A_2^2}$ applies to sinusoids in quadrature ($\frac\pi2$ phase difference), such as when summing a sine and a cosine.

Signals in phase just add up their amplitudes, $A_1+A_2$, while signals in opposition subtract, $|A_1-A_2|$.

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  • $\begingroup$ I've been staring at this for hours yesterday, and then this morning after I posted the question and suddenly realized my error. $\endgroup$ – rhody Jul 13 '18 at 18:26
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You wrote:

$$A_1 \sin (\omega t + \theta_1) + A_2 \sin (\omega t + \theta_2) = A_3 \sin (\omega t + \theta_3)$$

The Wolfram article you cite says:

It is always possible to write a sum of sinusoidal functions $$f(\theta)=acos(\theta)+bsin(\theta )$$
as a single sinusoid the form $$f(\theta)=ccos(\theta+\delta)$$ $$ A_3 = \sqrt{A_1^2 + A_2^2} $$

So you have sine plus sine, while Wolfram has sine plus cosine. To convert Wolfram's cosine to a sine, you need to shift the phase by $\frac{\pi}2$, which then makes the cosine of the phase difference equal to zero.

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Waves with no phase difference (or even pi's) directly add up their amplitudes to form a new wave.

$$A_1 \sin (\omega t) + A_2 \sin (\omega t) = (A_1+A_2) \sin (\omega t )$$

The $A_3$ you prescribed is for waves with phase difference $(\theta_1 - \theta_2)=\frac{\pi}{2}$.

The equation you got putting $\theta_1=\theta_2=0$ is correct and simplifies to $A_3=(A_1+A_2)$.

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The result given in Equation (21) of Wolfram site

for adding in-phase waves of same frequency with zero relative phase angle gives

$$A_3= \sqrt{A_1^2+A_2^2+ 2 A_1 A _2}= \pm (A_1+A_2) $$

The other result came about when the phase difference is $\pi/2$ obtained directly by Pythagoras thm as diagonal length of a rectangle of vectors or phasors.

The parallelogram of amplitudes can be drawn and the situation becomes clear. There are three situations. In-phase,add; Out of phase, subtract, or take the difference; when relative phase difference is $\pi/2$ take root square sum.

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