Is it possible to take the correlation coefficent of two continuous equations? I know that in general, the equation for calculating the Pearson correlation coefficient is given here (from wikipedia)
However, this seems to only apply for a given sample of points. Is it possible (or even a mathematically sound proposition) to calculate the correlation coefficient between equations between some bounds?
Say, for example, $$ y_1 = \sin(x), y_2 = \sin(x+45)$$ 
In python, I can generate discrete values for these equations at $1000$ points between $0$ and $360$ degrees, and calculate the correlation coefficients between the two resulting sets that way. But is there some way to do this with just the initial equations?
 A: If I understand correctly, $X$ is a uniformly-distributed random variable on the interval $[0,2\pi]$. (I'm changing from degrees to radians, hence the $2\pi$ instead of $360$ degrees.)
Assuming you're telling Python to generate random numbers between $0$ and $2\pi$, this is exactly what you'll get - a sample from a uniform distribution on $[0, 2\pi]$.
The random variables $Y_1$ and $Y_2$ are then defined as $Y_1 = \sin X$ and $Y_2 = \sin(X+\frac \pi 4)$.
We can then compute:
$$ E[Y_1 Y_2] = \frac 1 {2\pi} \int_0^{2\pi} \sin x \ \sin (x+\tfrac \pi 4) \ dx = \frac 1{2 \sqrt 2}.$$
$$ E[(Y_1)^2] = \frac 1 {2\pi} \int_0^{2\pi} \sin^2 x  \ dx = \frac 1 2.$$
$$ E[(Y_2)^2] = \frac 1 {2\pi} \int_0^{2\pi} \sin^2 (x+\tfrac \pi 4)  \ dx= \frac 1 2.$$
and
$$ E[Y_1] = \frac 1 {2\pi} \int_0^{2\pi} \sin x  \ dx = 0.$$
$$ E[Y_2] = \frac 1 {2\pi} \int_0^{2\pi} \sin  (x+\tfrac \pi 4) \ dx = 0.$$
Hence
$$ Cov[Y_1, Y_2] = E(Y_1 Y_2) - E(Y_1)E(Y_2) = \frac 1{2\sqrt 2}$$
$$ Var[Y_1] = E((Y_1)^2) - E(Y_1)^2= \frac 1 2$$
$$ Var[Y_2] = E((Y_2)^2) - E(Y_2)^2= \frac 1 2$$
So the correlation coefficient is
$$ \rho_{Y_1, Y_2} = \frac{Cov(Y_1, Y_2)}{\sqrt{Var(Y_1)Var(Y_2)}} = \frac 1 {\sqrt 2}.$$
By the way, if you change the "offset angle" of $\frac \pi 4$ (i.e. $45$ degrees) to an arbitrary angle $\alpha$, then the correlation coefficient will be $\cos \alpha$. You can prove this using the same method.
P.S. Just for fun, why don't you see what this code does?

