Entry 30

There's this unusual trigonometric relation. If,$$\sin(x+y) = 2\sin\big(\tfrac{1}{2}(x-y)\big)$$$$\sin(y+z) = 2\sin\big(\tfrac{1}{2}(y-z)\big)$$then$$\big(\tfrac{1}{2}\sin x\cos z\big)^{1/4}+\big(\tfrac{1}{2}\cos x\sin z\big)^{1/4} =\big(\sin 2y)^{1/12}$$The example by Ramanujan was,$$\begin{aligned}x &= \frac{\pi-\arcsin(p)}{2}=1.094\dots,\quad p=(4+\sqrt{15})^2\,\phi^{-9}\\ y &=\frac{\arcsin\big(\phi^{-3}\big)}{2}=0.119\dots\\z &=\frac{\arcsin(q)}{2}=0.0001\dots,\quad q=(4-\sqrt{15})^2\,\phi^{-9}\end{aligned}$$with golden ratio \(\phi=\frac{1+\sqrt{5}}{2}\). This implies the remarkable identity in radicals$$\Big(\tfrac{(\sqrt{1+p}\,+\sqrt{1-p})\,(\sqrt{1+q}\,+\sqrt{1-q})}{2^3}\Big)^{1/4}+\Big(\tfrac{(\sqrt{1+p}\,-\sqrt{1-p})\,(\sqrt{1+q}\,-\sqrt{1-q})}{2^3}\Big)^{1/4}=\phi^{-1/4}$$ with \(p=(4+\sqrt{15})^2\,\phi^{-9}\), and \(q=(4-\sqrt{15})^2\,\phi^{-9}\) as defined above. This was Question 359 of JIMS 4, p.78. (See The Problems Submitted by Ramanujan to the Journal of the Indian Mathematical Society, p. 9, by Bruce Berndt, et al.)