Derivation of Phase Angle Addition Formula

[kataya]

For two phasors $$a$$ and $$b$$, with magnitudes of A,B respectively and phase angles of $$\phi_{a)$$ and $$\phi_{b}$$, the angle of the sum of the phasors (call it p) has a magnitude of:

$$P^{2} = A^{2} + B^{2}$$ ,

and a phase angle of:

$$\phi_{p} = -tan^{-1}(\frac{B}{A})$$

The magnitude identity makes sense, as it is the geometric sum of the two vectors, but I am having trouble deriving the phase angle identity.

My original idea was to have something like this:

$$\phi_{p} = tan^{-1}(\frac{A^{2}sin^{2}\phi_{a} + B^{2}sin^{2}\phi_{b}}{A^{2}cos^{2}\phi_{a} + B^{2}cos^{2}\phi_{b}})$$

But I could not get anywhere with that, or at the very least work it into the form given above.

Does anyone know how to derive the angle formula? Many thanks

 

[CEL]

The formulas for magnitude and phase you gave are true only if the two phasors are orthogonal.