Rewriting solutions in the standard form: simple harmonic motion The differential equation: $y''+\omega^2 y=0$ has as a general solution: $$y=A\cos{(\omega t)}+B\sin{(\omega t)}$$
By taking:
$$A=R\cos{(\omega t_0)}$$ and $$B=R\sin{(\omega t_0)}$$
We can rewrite the general solution into: $$y=R\cos{(\omega(t-t_0))}$$
However, this is also a solution:
$$y=\alpha\cos{(\omega(t-t_0))+\beta\sin{(\omega(t-t_0))}}$$
QUESTION: 
I understand we can rewrite this last solution, $y=\alpha\cos{(\omega(t-t_0))+\beta\sin{(\omega(t-t_0))}}$ in the form of $y=A\cos{(\omega t)}+B\sin{(\omega t)}$ (using some trig identities)
But... Can we also rewrite $y=\alpha\cos{(\omega(t-t_0))+\beta\sin{(\omega(t-t_0))}}$ in the form of $y=R\cos{(\omega(t-t_0))}$? If so, how?
 A: Let $f(t)=\alpha\cos\omega(t-t_0)+\beta\sin\omega(t-t_0)$. Then as usual we have $f(t)=\alpha\cos\omega t_0\cos\omega t-\alpha\sin\omega t_0\sin\omega t+\beta\sin\omega t_0\cos\omega t+\beta\cos\omega t_0\sin\omega t$ $=(\alpha\cos\omega t_0+\beta\sin\omega t_0)\cos\omega t-(\alpha\sin\omega t_0-\beta\cos\omega t_0)\sin\omega t$ $=\gamma\cos\omega t-\delta\sin\omega t$, where $\gamma=\alpha\cos\omega t_0+\beta\sin\omega t_0,\delta=\alpha\sin\omega t_0-\beta\cos\omega t_0$ are constants.
Now we use the reverse procedure. Put $h=\frac{\gamma}{\sqrt{\gamma^2+\delta^2}},k=\frac{\delta}{\sqrt{\gamma^2+\delta^2}}$ and take $t_1$, so that $h=\cos\omega t_1,k=\sin\omega t_1$. Let $R=\sqrt{\gamma^2+\delta^2}$. Then we have $f(t)=R(\cos\omega t\cos\omega t_1-\sin\omega t\sin\omega t_1)=R\cos\omega(t-t_1)$.
A: We can also do it directly:
$$y=\alpha\cos(\omega(t-t_0))+\beta\sin(\omega(t-t_0))$$
Take:
$$\alpha=R\cos{\omega t_k}$$
$$\beta=R\sin{\omega t_k}$$
Where:
$$R=\sqrt{\alpha^2+\beta^2}$$
$$\omega t_k=\arctan\frac{\beta}{\alpha}$$
Then:
$$y=R\cos(\omega(t-t_0-t_k))$$
Take:
$$t_1=t_0+t_k$$
Then:
$$y=R\cos(\omega(t-t_1))$$
