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Friday, September 30, 2016

Entry 30

There's this unusual trigonometric relation. If,sin(x+y)=2sin(12(xy))sin(y+z)=2sin(12(yz))then(12sinxcosz)1/4+(12cosxsinz)1/4=(sin2y)1/12The example by Ramanujan was,x=πarcsin(p)2=1.094,p=(4+15)2ϕ9y=arcsin(ϕ3)2=0.119z=arcsin(q)2=0.0001,q=(415)2ϕ9with golden ratio ϕ=1+52. This implies the remarkable identity in radicals((1+p+1p)(1+q+1q)23)1/4+((1+p1p)(1+q1q)23)1/4=ϕ1/4 with p=(4+15)2ϕ9, and q=(415)2ϕ9 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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