I am solving a simple dynamics problem in a plane ruled by the equation $ m\ddot{\vec{r}}=-k\vec r$ where $k \gt 0$. I have managed to show that the general solution to the equation is
$$\vec r(t) = \vec A\space \sin\space\omega t + \vec B\space \cos\space\omega t$$ where $\vec A$ and $\vec B$ are constant vectors, $\omega = \sqrt{\frac k m}$.
I would now need to show that this general solution may be rewritten as
$$\vec r(t) = \vec a\space \sin(\omega t + \phi) + \vec b\space \cos(\omega t + \phi)$$
where $\vec a$ and $\vec b$ are constant orthogonal vectors and $\phi$ is a constant phase.
My attempts:
Starting with what I am supposed to arrive to, I noticed that the expression for $\vec A$ and $\vec B$ in terms of $\vec a$ and $\vec b$ is very similar to the expression for coordinate rotation in 2D. But that would imply that the general solution itself must have had two orthogonal vectors and the final result would be trivial.
By choosing coordinate axes such that the line $y=x$ is identical with the axis of the angle between $\vec A$ and $\vec B$ and considering projections of $\vec A$ and $\vec B$ on these axes, one could in theory derive the expression mentioned above and collapse it to the desired result, though this is a very tricky argument and would not probably pass as a proof.
Finding an orthogonal pair using the Gramm-Schmidt method. To no avail.
Expressing $\vec A$ and $\vec B$ coordinate-wise and using trigonometric formulas leads to pleasant expressions except for that it is impossible to show that the variable "representing phase" for $\vec a$ is the same as for $\vec b$, i.o.w. that the phase/rotation for time $t = 0$ is the same.
Am I overlooking something simple? I do not believe that this problem should be taking up this much time.
