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).
Best Answer
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, $|A_1-A_2|$.