TITLE: Linear combinations of sine and cosine
QUESTION [10 upvotes]: If you take a linear combination of the cosine and sine function, then the result is again a sinusoid, but with a new amplitude and phase shift.
$$a \cos(\theta) + b \sin(\theta) = A \cos(\theta + \theta_0)$$
This is one of those facts which I've always felt should be more obvious to me than it is. I can't help feeling that I might be missing out on some good "conceptual reason" for this to be true. I did a bit of thinking about it today and I came up with the following explanation which I'm not particularly satisfied with. I feel like there should be some perspective on this which renders the fact obvious. 

Multiplying the polar equation $r = 2a \cos(\theta) + 2b \sin(\theta)$ by $r$ yields $x^2+y^2 = 2a x + 2by$. Rearranging terms and adding $a^2 + b^2$ to both sides gives $x^2 - 2ax + a^2 +y^2 -2by + b^2 = a^2 + b^2$ which is identical to
  $$(x-a)^2 + (y-b)^2 = a^2 + b^2$$
  which we recognize as the equation for the circle with centre at $(a,b)$ and passing through the origin. This hints that $r = 2a \cos(\theta) + 2b \sin(\theta)$ should define a parametrization of this circle and, indeed, this holds. This circle can be rotated about the origin so as to have its centre on the $x$-axis. Such a circle will have the simpler equation
  $$r = D \cos(\theta) $$
  where $D=2(a^2+b^2)$ is the diameter of the circle. In polar coordinates, performing this rotation amounts to introducing a phase $\theta_0$. In fact $\theta_0$ is an argument for the point $(a,b)$. 

Note I'm not really looking for rigorous proofs of this fact, more like interpretations, really.

REPLY [3 votes]: The complex representation (or rotational matrix representation) gives a nice way of visualizing without actually saying we are using complex numbers. Cosine and sine are position of a point that started at the x axis and was rotated by angle $\theta$. $a\cos\theta$ is the segment on the $x$ axis at distance $a$, while $b\sin\theta$ has to start with a phase shift of quarter circle. Adding them means fixing the second circle to the point of the first circle.
Because both rotate by the same angle, we can just add (a,0)+(0,-b) in unrotated frame and rotate both by $\theta$. The resulting $x$ segment is your cosine with a phase shift.

This method also works if the LHS already contains phase shifts.