Addition formula for $\text{sn}(u)=\text{sn}(u,k)$

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I have just learned the definition of the first Jacobian elliptic function $\text{sn}(u)=\text{sn}(u,k)$, defined as the sine of the inverse function of $$F(\phi,k):=\int_0^\phi \frac{dx}{\sqrt{1-k^2\sin^2(x)}}$$ with respect to $\phi$; that is, if $u=F(\phi,k)$, then $\text{sn}(u,k)=\sin\phi$. I also know the analogous definitions of $\text{cn}(u,k)$ and $\text{dn}(u,k)$.

QUESTION. I have seen the following addition formula and would like to know how to prove it: $$\text{sn}(u+v)=\frac{\text{sn}(u)\text{cn}(v)\text{dn}(v)+\text{sn}(v)\text{cn}(u)\text{dn}(u)}{1-k^2\text{sn}^2(u)\text{sn}^2(v)}$$ How is this formula proven? Does it require any "advanced machinery" that a beginner would not be familiar with, or is it a simple proof?