What is the significance of these two functions' relationship? I noticed that the graph of $x^2$ looked somewhat like $\sin(x)$, so I tried to "recreate" $\sin(x)$ using it.
$-(-1)^{\lfloor x/2\rfloor}\cdot(x-2\cdot\lfloor x/2\rfloor-1)^2 + (-1)^{\lfloor x/2\rfloor}$
Anyway, that equation came from me trying to take the lower parts of $x^2$ (between -1 and 1) and squish them together, alternating positive and negative. Graph it and you'll see what I mean.
Anyway, it didn't quite match up with $\sin(x)$, which wasn't surprising. The wavelength was all wrong. But if I do $\sin(\frac{x\pi}{2})$, the wavelength matches perfectly.
To my dismay, the functions are not identical. $\sin(x)$ is always a bit closer to zero (except on the turning/inflection points).
Still, it's interesting. But is there any inherent relationship here or am I just throwing things together all haphazardly? I feel like something's important because of the $\pi/2$, but then again it might just be something obvious and silly I'm overlooking.
Any insights would be terrific. Thank you!
 A: Well, the functions definitely aren't the same. 
Your floor function has period $4$ - it alternates between the upward and downward parabolas. $\sin(x)$ has a period of $2 \pi$. So naturally, when you scale the argument of the $\sin(x)$ to $\sin(\pi x/2)$, you've reduced the period by a factor of $\pi/2$. So $2 \pi/(\pi/2) = 4$. The period matches.
The sine function kind of looks like a parabola because of its concavity, but it is not one. You might, however, be interested in Taylor series expansions, which approximate functions with polynomials to arbitrary accuracy. (Note, though, that's different from what you're doing.)
A: What explains the near-parabolic shape of the sine function are Taylor Series.  You'd need some Calculus to fully understand the Taylor Series, but here it is (in radians):
$$\sin(x)=\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$$
In summation notation:
$$\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$

Each new line represents more and more parts of the summation.  The red line is $y=\frac{x}{1!}$, the orange is $y=\frac{x}{1!}-\frac{x^3}{3!}$, and so on $\dots$
If you take a closer look at the expanded form of the sine function, you will notice two things.  The denominator is getting increasingly large while the exponent in the numerator gets increasingly large.
Now, let's assume $x$ is very small, in fact, we will imagine it is less than $1$ for simplicity.  You will notice that $x>x^3>x^5>x^7>\dots$ if $x<1$.
You will also notice that the denominator makes each term in the sequence increasingly small.
From all of this, we get a much simpler approximation:
$$\sin(x)=\frac{x}{1!}-\frac{x^3}{3!}+O(x^5)$$
The $O(f(x))$ means that any error you encounter should be small and is due to something with a growth rate of $f(x)$.  Since $x^5$ is really small, we can pretty much ignore it.  Everything after $x^5$ is even smaller, so we can ignore those as well.
So what you observed is that the sine function has an approximately polynomial growth for small values of $x$.  We can extend this to any $x$, but best to keep it small.
You got close using your $x^2$ sort of polynomial, but it obviously fails.  As with the approximate Taylor series, yours works for a certain region of $x$ and error encountered is due to some amount in another term of the Taylor series.
And interesting trick you pulled off with the floor function.
You may note that your function has $x^2$, whereas sine doesn't.  However, if we shift over into cosine, cosine does indeed have an $x^2$ term.
$$\cos(x)=\frac{x^0}{0!}-\frac{x^2}{2!}+O(x^4)=1-\frac{x^2}2+O(x^4)$$
The error you experience is due to $x^4$.
Looking at your formula, I deduce that it is more simplified as
$(-1)^{\lfloor x/2\rfloor}(1-f(x))$
For now, let us not concern ourselves with the exact formula of $f(x)$, but rather we should notice the growth rate of it.  It is, interestingly, having a growth rate of $x^2$.  So, it is approxiamately $(1-x^2)$, neglecting the oscillating $(-1)$.
So, if you want to improve this, I would do one simple change, use $1-\frac{f(x)}2$ instead, it is closer to the Taylor Series of cosine.  Try it, it works better.  (It also explains why sine was always a little bit closer to $0$)
