trigonometry of interfering waves I have a book which makes the following claim:
$$ A \cos( \omega t + \phi ) = A_1 \cos( \omega t + \phi_1 ) + A_2 \cos( \omega t + \phi_2 ) $$
...where:
$$ A^2 = A_1^2 + A_2^2 + 2 A_1 A_2 \cos( \phi_1 - \phi_2 ) $$
...and:
$$ \tan( \phi ) = \frac{ A_1 \sin( \phi_1 ) + A_2 \sin( \phi_2 ) }{ A_1 \cos( \phi_1 ) + A_2 \cos( \phi_2 ) } $$
However, I've been unable to derive this set of relations, and I've used every trig identity I've been able to think of.  Either I'm missing something, or there's a simplifying assumption or approximation the author is making that they aren't explicitly stating.
If it helps at all, this is a book about optics (specifically: holography), and these formulas are being used to describe the interference of spherical wavefronts of light from two point sources.  The original sources $O_1$ and $O_2$ have phases $\phi_1$ and $\phi_2$, respectively, and at the point of measurement $M$ the phase of the resulting wavefront is $\phi$ (and $\omega$ is the same throughout).  The amplitues of the waves at $M$, from $O_1$, and from $O_2$ are $A$, $A_1$, and $A_2$, respectively.
Additionally, we have: $\omega = 2 \pi \nu$ and $k = 2 \pi / \lambda$, where $\nu$ is frequency and $\lambda$ is wavelength; and we also have $ \phi_1 = \phi_{01} - k r_1 $ and $ \phi_2 = \phi_{02} - k r_2 $, where $r_1$ and $r_2$ are the distances to $M$ from $O_1$ and $O_2$, respectively, and $\phi_{01}$ and $\phi_{02}$ are the initial phases at those same sources.  Though I'm not sure how relevant this last group of quantities is to the part of the problem I'm stuck on.
Am I trying too hard to find an exact solution?  Is this really just an approximation?
I'd like to post photographs of the relevant pages in the book, but I'm not sure whether that complies with site policy (or more importantly, with copyright law - does it constitute fair-use?).  The book is "Holography and its Application", by Yu. I. Ostrovsky; English edition published in 1977, Russian edition published in 1973.  (I'd actually be surprised if it were still in-print, and/or any publishers still hold the rights to it, but I'm not taking any chances.)
Also, this is self-study, it's not for any class.
 A: Okay, let's have a go at this by "brute force". Writing $\theta=\omega t+\phi$, and for the time being $\phi_1-\phi=a$, $\phi_2-\phi=b$, we expand the right-hand side as
$$ A_1 \cos{(\theta+a)}+A_2 \cos{\theta+b} = \cos{\theta}(A_1\cos{a}+A_2\cos{b}) - \sin{\theta}(A_1\sin{a}+A_2\sin{b}) $$
Now somehow we want the second term to be zero and the first to have coefficient $A$. So
$$ 0 = A_1 \sin{(\phi_1-\phi)}+ A_2 \sin{(\phi_2-\phi)} \\
= \cos{\phi}(A_1 \sin{\phi_1}+A_2\sin{\phi_2} ) -\sin{\phi}(A_1 \cos{\phi_1} + A_2 \cos{\phi_2} ), $$
which we can see turns into the $\tan{\phi}$ relationship.
Now, we have on the other hand
$$ A = A_1 \cos{(\phi_1-\phi)}+A_2\cos{(\phi_2-\phi)}. $$
Probably the sensible thing to do at this point is to start drawing pictures and find out why the thing we want looks so like the cosine rule, but let's try and push on with the algebra. Blindly expanding out, we have
$$ A = \cos{\phi} ( A_1 \cos{\phi_1}+A_2 \cos{\phi_2}) - \sin{\phi} (A_1 \sin{\phi_1} + A_2 \sin{\phi_2}). $$
Well now, these look familiar. The trick now is to get rid of $\phi$ as expediently as possible. Swapping the $\sin{\phi}$, we have
$$ A \sec{\phi}= \frac{(A_1 \cos{\phi_1}+A_2\cos{\phi_2})^2 + (A_1 \sin{\phi_1}+A_2\sin{\phi_2} )^2}{A_1\cos{\phi_1}+A_2\cos{\phi_2}}\\
= \frac{A_1^2+A_2^2+2A_1A_2(\cos{\phi_1}\cos{\phi_2}+\sin{\phi_1}\sin{\phi_2})}{A_1\cos{\phi_1}+A_2\cos{\phi_2}}
$$
Okay, now that's suspicious: we've ended up disappointingly close, since the numerator is what we want $A^2$ to be Let's call it $A'^2$. Now let's try squaring everything.
$$ A^2 \sec^2{\phi} = \frac{A'^4}{(A_1\cos{\phi_1}+A_2\cos{\phi_2})^2} $$
Now, $\sec^2{\phi}=1+\tan^2{\phi}$, and doing the suspiciously familiar calculation gives
$$ \sec^2{\phi} = \frac{A'^2}{(A_1\cos{\phi_1}+A_2\cos{\phi_2})^2} $$,
and hence
$$ A^2 A'^2 = A'^4, $$
so $A^2=A'^2$ as desired. There's bound to be a neater way of doing that last bit, and @rogerl's link looks like it has the geometrical method that seems the sensible way to go, but there's your answer.
A: $$
A \cos( \omega t + \varphi ) = A_1 \cos( \omega t + \varphi_1 ) + A_2 \cos(\omega t + \varphi_2 )
$$
Notice that if you know $\varphi_1$ and $\varphi_2$ and $A_1$ and $A_2$, you can find $\varphi$, but if you're given $\varphi$ you cannot find those other four quantities; the function that maps that quadruple to $\varphi$ is not one-to-one.  So we will therefore start with the right side of the identity and work toward the left side.
Letting $s=t+\dfrac{\varphi_1}\omega$, we have
\begin{align}
& A_1 \cos( \omega t + \varphi_1 ) + A_2 \cos( \omega t + \varphi_2 ) \\[6pt]
= {} & A_1\cos(\omega s) + A_2\cos(\omega s+(\varphi_1-\varphi_2)) \\[6pt]
= {} & A_1\cos(\omega s) + A_2\cos(\omega s+\varphi_3) \\[6pt]
= {} & A_1\cos(\omega s) + A_2\cos(\omega s)\sin\varphi_3+A_2\sin(\omega s)\cos\varphi_3 \\[6pt]
= {} & \underbrace{(A_1 + A_2\sin\varphi_3)}\cos(\omega s) + A_2\sin(\omega s) \\[6pt]
= {} & B_1\cos(\omega s) + B_2\sin(\omega s) \\[6pt]
= {} & B_3 \sin(\omega s+\chi)\text{ (by the conventional identity)}
\end{align}
where $B_3^2=B_1^2+B_2^2$ and $\tan\chi=B_1/B_2$.
In other words, we've reduced it to the more conventional identity.  Now you have to convert it back to an expression in $t$ and $\varphi$ instead of $s$ and $\chi$.
If there are questions about proving the more conventional identity, I could add a postscript later.
postscript:  OK, I'm back, so let's look at what I called "the more conventional identity":
$$
A\cos x + B\sin x = C\sin(x+\varphi), \text{ where }\tan\varphi = \frac A B \text{ and } C^2 = A^2+B^2.
$$
We have
\begin{align}
A\cos x+B\sin x & = C\left(\frac A C \cos x + \frac B C \sin x \right) \\[6pt]
& = C( \sin\varphi\cos x + \cos\varphi\sin x) \tag 1 \\[6pt]
& = C\sin(x+\varphi).
\end{align}
The justification of what we did at $(1)$ is that since $C=\sqrt{A^2+B^2}$, the sum of the squares of $A/C$ and $B/C$ is $1$; hence $(A/C,\,B/C)$ is a point on the unit circle; hence there is some angle $\varphi$ for which that pair is $(\sin\varphi,\cos\varphi)$.  Then finally we have
$$
\tan\varphi = \frac{\sin\varphi}{\cos\varphi} = \frac{A/C}{B/C} = \frac A B.
$$
A: Easiest is to consider two vectors of length $ A_1,A_2$ rotating with same angular speed. The resultant diagonal sum is obtained by Parallelogram Law. The argument also rotates as you have given at the same angular speed. 
A: Okay, let's start with the first equation:
$$ A \cos( \omega t + \phi ) = A_1 \cos( \omega t + \phi_1 ) + A_2 \cos( \omega t + \phi_2 ) \tag 1$$
We can substitute $ \omega t = \omega t' + \pi / 2 $ into $(1)$, and then switch from the dummy variable $ t' $ back to the original variable $t$, to show that the following is also true:
$$ A \sin( \omega t + \phi ) = A_1 \sin( \omega t + \phi_1 ) + A_2 \sin( \omega t + \phi_2 ) \tag 2$$
Taken together, both $(1)$ and $(2)$ are the real and imaginary parts of the following:
$$ A \exp( i\omega t + i\phi ) = A_1 \exp( i\omega t + i\phi_1 ) + A_2 \exp( i\omega t + i\phi_2 ) \tag 3$$
Factoring-out $\exp(i\omega t)$ leaves us with:
$$ A \exp( i\phi ) = A_1 \exp( i\phi_1 ) + A_2 \exp( i\phi_2 ) \tag 4$$
Which separates into real and imaginary parts:
$$ A \cos( \phi ) = A_1 \cos( \phi_1 ) + A_2 \cos( \phi_2 ) \tag 5$$
$$ A \sin( \phi ) = A_1 \sin( \phi_1 ) + A_2 \sin( \phi_2 ) \tag 6$$
Dividing $(6)$ by $(5)$ gives us:
$$ \tan( \phi ) = \frac{ A_1 \sin( \phi_1 ) + A_2 \sin( \phi_2 ) }{ A_1 \cos( \phi_1 ) + A_2 \cos( \phi_2 ) }$$
And taking the modulus-squared of both sides of $(4)$ gives us:
$$ A^2 = A_1^2 + A_2^2 + A_1 A_2 \exp( i\phi_1 - i\phi_2 ) + A_1 A_2 \exp( i\phi_2 - i\phi_1 ) $$
$$ = A_1^2 + A_2^2 + A_1 A_2 \exp( i\phi_1 - i\phi_2 ) + A_1 A_2 \exp( - i\phi_1 + i\phi_2 ) $$
$$ = A_1^2 + A_2^2 + A_1 A_2 \exp( +i ( \phi_1 - i\phi_2 ) ) + A_1 A_2 \exp( - i ( \phi_1 - i\phi_2 ) ) $$
$$ = A_1^2 + A_2^2 + 2 A_1 A_2 \cos( \phi_1 - \phi_2 ) $$
(Where a form of Euler's formula was used in that last step: $ 2 \cos(x) = \exp( +ix ) + \exp( -ix) $)
