Problem:
With all angles measured in degrees, the product ∏k=145csc2(2k−1)∘=mn, where m and n are integers greater than 1. Find m+n.
Solution:
Let
P=sin1∘sin3∘sin5∘⋯sin89∘,
and let
Q=sin2∘sin4∘sin6∘⋯sin88∘.
Then
PQ=sin1∘sin2∘sin3∘⋯sin89∘.
But also
PQ=sin89∘sin88∘sin87∘⋯sin1∘.
Therefore
P2Q2=sin1∘sin89∘sin2∘sin88∘sin3∘sin87∘⋯sin89∘sin1∘=sin1∘cos1∘sin2∘cos2∘sin3∘cos3∘⋯sin89∘cos89∘
Then
289P2Q2=(2sin1∘cos1∘)(2sin2∘cos2∘)(2sin3∘cos3∘)⋯(2sin89∘cos89∘)=sin2∘sin4∘sin6∘⋯sin178∘=(sin2∘sin4∘sin6∘⋯sin88∘)(sin92∘sin94∘sin96∘⋯sin178∘)=(sin2∘sin4∘sin6∘⋯sin88∘)(sin88∘sin86∘sin84∘⋯sin2∘)=Q2
Thus 289P2Q2=Q2. Because Q2=0, the requested product of cosecants equals P21=289. Because 89 is prime, this representation with integers greater than 1 is unique. The requested sum is 2+89=91.
The problems on this page are the property of the MAA's American Mathematics Competitions