There's this unusual trigonometric relation. If,sin(x+y)=2sin(12(x−y))sin(y+z)=2sin(12(y−z))then(12sinxcosz)1/4+(12cosxsinz)1/4=(sin2y)1/12The example by Ramanujan was,x=π−arcsin(p)2=1.094…,p=(4+√15)2ϕ−9y=arcsin(ϕ−3)2=0.119…z=arcsin(q)2=0.0001…,q=(4−√15)2ϕ−9with golden ratio ϕ=1+√52. This implies the remarkable identity in radicals((√1+p+√1−p)(√1+q+√1−q)23)1/4+((√1+p−√1−p)(√1+q−√1−q)23)1/4=ϕ−1/4 with p=(4+√15)2ϕ−9, and q=(4−√15)2ϕ−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.)
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