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.