In JIMS 4, p.78, Question 359 was asked by Ramanujan. (See *The Problems Submitted by Ramanujan to the Journal of the Indian Mathematical Society*, p. 9, by Bruce Berndt, et al.) If,

$$\sin(x+y) = 2\sin\big(\tfrac{1}{2}(x-y)\big)\tag1$$

$$\sin(y+z) = 2\sin\big(\tfrac{1}{2}(y-z)\big)\tag2$$

prove that,

$$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2y)^{1/12}\tag3$$

The example by Ramanujan was,

$$\begin{aligned} x &= \frac{\pi-\arcsin\big((\sqrt{5}-2)^3(4+\sqrt{15})^2\big)}{2}=1.094\dots\\ y &=\frac{\arcsin\big(\sqrt{5}-2\big)}{2}=0.119\dots\\ z &=\frac{\arcsin\big((\sqrt{5}-2)^3(4-\sqrt{15})^2\big)}{2}=0.0001\dots \end{aligned}$$

Ten years later, a 3-page proof was given in JIMS 15, p.114-117.

I got an email asking if there was a shorter proof. Considering the problem, I observed the following. Given the quartic,

$$a^3w^4+(1-3a^2)w^3+3a(1-a^2)w^2+a^2(3-a^2)w-a=0\tag4$$
with $a=\tan(\color{blue}y/4)$. Define,
$$x=4\tan^{-1} u\\z=4\tan^{-1} v$$
where $u,v$ are *two* appropriate roots of the quartic, then we get the ** same** bizarre relation as Ramanujan,

$$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2\color{blue}y)^{1/12}\tag5$$

Equivalently, those two roots $u,v$ obey,

$$\left(\frac{1}{2}\,\frac{4(u-u^3)}{(u^2+1)^2}\frac{v^4-6v^2+1}{(v^2+1)^2} \right)^{1/4}+\left(\frac{1}{2}\,\frac{4(v-v^3)}{(v^2+1)^2}\frac{u^4-6u^2+1}{(u^2+1)^2} \right)^{1/4}=\big(\sin 2\color{blue}y\big)^{1/12}\tag6$$

For example, let $y=1$, so $a=\tan(1/4)$, then $u,v$ are the two real roots of (4).

** Question**: Anyone knows a short proof for (3) and (6)?