How do I prove this curve has the shape of a logarithmic spiral? I am working with $\alpha(t) = \cfrac{e^{at}}{a^2+1}(\sin(t) + a\cos(t),a \sin(t) - \cos(t))$, a reparametrization of $\alpha(s) = \left(\cfrac{(as+b)\left(\sin\left(\cfrac{\log(as+b)}{a}\right) + a\cos\left(\cfrac{\log(as+b)}{a}\right) \right) }{a^2 + 1},\cfrac{(as+b)\left(a\sin\left(\cfrac{\log(as+b)}{a}\right) -
 \cos\left(\cfrac{\log(as+b)}{a}\right) \right)}{a^2 + 1} \right)$
the latter is parameterized by arclength, but obviously the former is easier to work with. Here, both $a$ and $b$ are constants. I'm trying to get it into something like $\beta(t) = e^t(\cos(t),\sin(t))$, but it's been tough.
 A: Define $\delta$ such that $\cot \delta = a$.  Then we have
$$
\sin t + a \cos t = \sin t + \cot \delta \cos t = \frac{1}{\sin \delta} (\sin \delta \sin t + \cos \delta \cos t) = \frac{\cos(t - \delta)}{\sin \delta}
$$
and
$$
a \sin t - \cos t = \cot \delta \sin t - \cos t = \frac{1}{\sin \delta} (\cos \delta \sin t - \sin \delta \cos t) = \frac{\sin(t - \delta)}{\sin \delta}
$$
Moreover, we have
$$
\sin \delta = \frac{1}{\csc \delta} = \frac{1}{\sqrt{1 + \cot^2 \delta}} = \frac{1}{\sqrt{1 + a^2}}
$$
and so putting this all together, we have
$$
\alpha(t) = \cfrac{e^{at}}{a^2+1}(\sin(t) + a\cos(t),a \sin(t) - \cos(t)) = \frac{e^{at}}{\sqrt{1 + a^2}} \left(\cos (t - \delta), \sin (t - \delta) \right),
$$
which is simply an exponential spiral that starts at an angle $\theta = -\delta$ when $t = 0$, rather than at $\theta = 0$.
In general, any phase-shifted sinusoid can be written in one of two ways: either $x(t) = C \cos (t + \delta)$, or $x(t) = A \sin t + B \cos t$.  Any expression in the first form can be rewritten in the second form, and vice versa.  If you're going to be spending a significant amount of time dealing with sinusoidal functions, it's a useful skill to be able to translate between these two forms, and I would encourage you to spend a bit of time exploring how to do this on your own.
A: Hint If we take a standard logarithmic spiral parameterization, $u \mapsto (R e^{c u} \cos u, R e^{c u} \sin u)$, and substitute $u := t + t_0$ and apply the angle sum identity, we get, e.g., for the first component,
$$R e^{c (t + t_0)} \cos (t + t_0) = e^{c t} (\color{red}{R e^{c t_0} \cos t_0} \cos t - \color{blue}{R e^{c t_0}\sin t_0} \sin t) .$$
We can thus try to match this with the first component
$$e^{at} \left(\color{\red}{\frac{a}{a^2 + 1}} \cos t + \color{blue}{\frac{1}{a^2 + 1}} \sin t\right)$$
of $\alpha(t)$ by choosing appropriate parameters $c, R, t_0$.

 Additional hint Comparing like coefficients immediately gives $c = a$, and taking the sum of the squares of the red and blue coefficients then gives $R^2 e^{2 a t_0} = \frac{1}{a^2 + 1}$, so $R = \frac{e^{-a t_0}}{\sqrt{a^2 + 1}}$. Finally, comparing the red and blue terms gives $$\cos t_0 = \frac{a}{\sqrt{a^2 + 1}}, \qquad \sin t_0 = -\frac{1}{\sqrt{a^2 + 1}},$$ and we can choose an angle $t_0$ that satisfies both of these equations---taking $t_0 := \operatorname{arccot} (-a)$ will do. Thus, if we can show that these choices also make the second components of our standard parameterization and the formula for $\alpha(t)$ agree, we'll have shown that $\alpha(t)$ is a parameterization of a logarithmic spiral.

