On Ramanujan's Question 359

Thanks to a deleted answer by G. Manco, we finally have a partial answer after four years to a closed-form solution $x,y,z$ to, $$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2y)^{1/12}\tag3$$ in terms of the Ramanujan G function. Using the Dedekind eta function $\eta(\tau)$, this can be calculated in Mathematica as, $$G_n=\frac{2^{-1/4}\,\eta^2(\sqrt{-n})}{\eta\big(\tfrac{\sqrt{-n}}{2}\big)\,\eta(2\sqrt{-n})}\tag4$$ However, because of the constraints $(1),(2)$ in the post, there is a difference when $1\leq n\leq9$ and $n\geq9$. Thus,

If $1\leq n\leq9$: $$\begin{aligned} x &=\pi/2-\tfrac12\arcsin G_{n/9}^{-12}\\ y &=\tfrac12\arcsin G_{n}^{-12}\\ z &=\tfrac12\arcsin G_{9n}^{-12} \end{aligned}$$ If $n\geq9$: $$\begin{aligned} x &=\tfrac12\arcsin G_{n/9}^{-12}\\ y &=\tfrac12\arcsin G_{n}^{-12}\\ z &=\tfrac12\arcsin G_{9n}^{-12} \end{aligned}$$ We recover Ramanujan's solution $x,y,z \approx 1.094,\; 0.119,\; 0.0001$ by plugging in $n=5$ to the first set of formulas. Also, since,

$$\sin(\tfrac12\arcsin \lambda)=\tfrac12\big(\sqrt{1+\lambda}-\sqrt{1-\lambda}\big)$$ $$\cos(\tfrac12\arcsin \lambda)=\tfrac12\big(\sqrt{1+\lambda}+\sqrt{1-\lambda}\big)$$ then $(3)$ can be expressed in radical form,

$$\left(\tfrac12\Big(\tfrac{\sqrt{1+\alpha}+\sqrt{1-\alpha}}2\Big)\Big(\tfrac{\sqrt{1+\beta}\,\color{blue}\pm\,\sqrt{1-\beta}}2\Big)\right)^{1/4}+\left(\tfrac12\Big(\tfrac{\sqrt{1+\alpha}-\sqrt{1-\alpha}}2\Big)\Big(\tfrac{\sqrt{1+\beta}\,\color{blue}\mp\,\sqrt{1-\beta}}2\Big)\right)^{1/4}=\gamma^{1/12}$$ where $\alpha = G_{n/9}^{-12},\;\beta= G_{9n}^{-12},\;\gamma= G_{n}^{-12}$ and the $+,-$ case for $n\leq9$ and $-,+$ case for $n\geq9$.

P.S. The second part of the question is still a bit tricky to prove.

An elegant solution to question 359 submitted by Ramanujan to Journal of the Indian Mathematical Society.

The three equations (1)-(3) are three modular equations: the first two are of third degree and the last one of ninth degree. These equations are in Ramanujan's notebooks: http://www.imsc.res.in/~rao/ramanujan/index.html.

If $\alpha$ be of the $1^{st}$, $\beta$ the $3^{rd}$ and $\gamma$ the $9^{th}$ degree then.

$\sqrt{\alpha(1-\beta)}$+$\sqrt{\beta(1-\alpha)}$=$2(\alpha\beta(1-\alpha)(1-\beta))^\frac {1} {8}$$\tag1$

$\sqrt{\gamma(1-\beta)}$+$\sqrt{\beta(1-\gamma)}$=$2(\beta \gamma (1-\beta)(1-\gamma))^\frac{1} {8}$$\tag2$

and

$(\alpha(1-\gamma))^\frac{1}{8}$+ $(\gamma(1-\alpha))^\frac{1}{8}$=$2^\frac{1}{ 3}$($\beta(1-\beta))^\frac{1}{24}$ $\tag3$

Here is my interpretation. The reading key is:

$\sqrt{\alpha}=\sin{x}$, $\sqrt{\beta}=\sin{y}$, and $\sqrt{\gamma}=\sin{z}$

moreover

$\sqrt{1-\alpha}=\cos x$, $\sqrt{1-\beta}=\cos y$ and $\sqrt{1-\gamma}=\cos z$.

and with class invariants

$4\alpha(1-\alpha)G_{\frac n 9}^{24}=1$, $4\beta(1-\beta)G_{n}^{24}=1$, and $4\gamma(1-\gamma)G_{9n}^{24}=1$

or

$\sin{2x}.G_{\frac {n} {9}}^{12}=1$ , $\sin{2y}.G_{n}^{12}=1$ , and $\sin{2z}.G_{9n}^{12}=1$.

For $n=5$ we have

$\sin 2x=2\sqrt{\alpha(1-\alpha)} =G_{\frac {5} {9}}^{-12}$=$((\sqrt 5+2)^{\frac 1 4}(4-\sqrt{15})^{\frac 1 6})^{-12}=(\sqrt 5-2)^3(4+\sqrt 15)^2$

$\sin 2y=2\sqrt{\beta(1-\beta)}=G_{5}^{-12}=\big((\frac {\sqrt 5+1}{2})^{\frac1 4}\big)^{-12}=(\sqrt 5-2)$

$\sin 2z=2\sqrt{\gamma(1-\gamma)}=G_{45}^{-12}=((\sqrt 5+2)^{\frac 1 4}(4+\sqrt{15})^{\frac 1 6})^{-12}=(\sqrt 5-2)^3(4-\sqrt 15)^2$

$(\frac 1 2 \sin x \cos z)^{\frac 1 4}=\frac{ 1} {\sqrt 2}(\sqrt 5-2)^{\frac 3 4}\big(\sqrt \frac {7+3\sqrt 5}{4}+\sqrt \frac {3+3\sqrt 5}{4}\big)$

$(\frac 1 2 \sin z \cos x)^{\frac 1 4}=\frac{ 1} {\sqrt 2}(\sqrt 5-2)^{\frac 3 4}\big(\sqrt \frac {7+3\sqrt 5}{4}-\sqrt \frac {3+3\sqrt 5}{4}\big)$

adding we have

$\frac{ 1} {\sqrt 2}(\sqrt 5-2)^{\frac 3 4}\big(2\sqrt \frac {7+3\sqrt 5}{4}\big)=(\sqrt 5-2)^{\frac 1 {12}}=(\frac {\sqrt 5-1}{2})^{\frac1 4}$

This result is also equal to

$(\sin 2y )^{\frac {1} {12}}=\frac {1} {G_{5}}=(\frac {\sqrt 5-1}{2})^{\frac{1} {4}}$

I calculated several solutions for $n=1,1/2,2,3,1/3,4,1/4,5,1/5,6,1/6,7,1/7,8,1/8,9,1/9,10,1/10,15,1/15.$

For $n=12$ we have

$\sin x$=$(\sqrt{2}+1)^2(\sqrt{3}-\sqrt{2})^2$

$\sin y$=$(\sqrt{2}-1)^2(\sqrt{3}-\sqrt{2})^2$

$\sin z$=$\frac{2-\sqrt{2+\sqrt{\delta}}} {2+\sqrt{2+\sqrt{\delta}}}$

with

$\delta=3-100.2^{\frac{1} {3}}+80.2^{\frac{2} {3}}$.