I need help, I'm looking at Elliptic integrals and why you can use AGM to compute it. I got the point why, but I didn't understand some steps where Gauss make a proof on it. He came with new variable $\varphi'$, where
$\sin(\varphi) = \frac{2a\sin^2(\varphi')}{(a+b)+(a-b)\sin^2(\varphi')}$,
then you need to make $\cos^2(\varphi) = 1 -\sin^2(\varphi) $, and that's where the struggle begins, so you got like
$\cos^2(\varphi) = \frac{(a+b)^2+2(a^2-b^2)\sin^2(\varphi')+(a-b)^2\sin^4(\varphi') - 4a^2\sin^4(\varphi')}{[(a+b)+(a-b)\sin^2(\varphi')]^2}\qquad(1)$
now you need to put in $ a_1 = \frac{a+b}{2}, b_1=\sqrt{a\cdot b} $ so you get
$\cos^2(\varphi) =\frac{4a_1^2-4(2a_1^2-b_1^2)\sin^2(\varphi')+4(a_1^2-b_1^2)\sin^4(\varphi')}{[(a+b)+(a-b)\sin^2(\varphi')]^2}=\frac{4a_1^2\cos^4(\varphi')+4b_1^2\sin^2(\varphi')\cos^2(\varphi')}{[(a+b)+(a-b)\sin^2(\varphi')]^2} \qquad (2)$
Soo, can be someone so kind and show me how I put in $a_1,b_1$, into $(1)$ to get $(2)$ please, I just don't see it and want to get it clear.
Thank you for you time and have a nice day.
The substitution is C&P'ed incorrectly. The correct one is $$\sin\varphi=\frac{2a\color{red}{\sin\varphi'}}{(a+b)+(a-b)\sin^2\varphi'}.$$ (Don't forget to take the $d\varphi$ part into account too.)