Circles and Sine Waves All examples I've come across use unit circles to relate with sine waves. How about circles that are not unit circles? Won't the amplitude of the sine wave not correspond to the vertical displacement of the line vector that is spinning in the circle? Since the radius/line vector is not 1 unit anymore?
 A: As you stated, we can express any point on a unit circle $$ x=\cos(\theta) \\ y=\sin(\theta)$$ for $\theta \in [0,2\pi]$.
There is an implicit factor of $1$ in front of the terms on the RHS corresponding to radius $r=1$ (hence unit circle).
So when the circle is not a unit circle but one with radius $r\gt 1$ then we would simply write the equations as $$x=r\cos(\theta) \\ y=r\sin(\theta)$$
These equations now correspond to points on a circle with a radius greater than one, and define circular motion for an object who's distance from the center using Pythagorean theorem is $$\sqrt{x^2 + y^2}=\sqrt{r^2\left(\cos^2(\theta)+\sin^2(\theta)\right)}=\sqrt r^2=r$$
A: We use the unit circle because sine and cosine are ratios of the $x$, $y$, and the radius of the circle. Those ratios (e.g. $\tan \theta =\tfrac{y}{x}$ or $\sin \theta = \tfrac{y}{1}$) do not depend on the radius. Another radius just scales up $x$ and $y$ and adds another variable $r$ because it’s not $1$ anymore.
Trig works with any sized circle. But $\cos \theta$ is independent of the circle used to find it.
A: Amplitude does not matter, it is taken as unity for a simple start or convenience.
In  differential equation descriptions it is totally absent, appearing only through chosen initial/boundary conditions as in the examples:
$$ y''(x) + \omega^2 y(x)=0 $$
or
$$ \cos \omega t \;y^{''}(t)) +\omega \sin\omega t  \;y'(x)=0 $$
There are $many$ equations in mathematical physics that represent oscillatory phenomenon where particular amplitude constants do not appear explicitly. eg $\Delta ^2 w =0 $..
