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

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?

• Maybe not with all the details you are asking for but pages 4,5,6,7,8 of this seem to be helpful. It looks like there is a result which is true of a broader class of functions and this is just an application of this. – Mason Nov 15 '18 at 22:05
• @Mason Wow, that's great! That is exactly what I was looking for. Thank you so much! – Frpzzd Nov 15 '18 at 22:09
• @Mason Well, if I can work out the details then I will post it as an answer to my own question. – Frpzzd Nov 15 '18 at 22:27
• This is a very standard result and is presented in most textbooks which cover elliptic functions. You may also have a look at my post based on Cayley's book. – Paramanand Singh Nov 21 '18 at 15:14
• In general no advanced machinery is needed to understand elliptic functions unless you want to study its link with algebraic number theory. For me this link has been so notoriously difficult. – Paramanand Singh Nov 21 '18 at 15:17