Problem:
With all angles measured in degrees, the product βk=145βcsc2(2kβ1)β=mn, where m and n are integers greater than 1. Find m+n.
Solution:
Let
P=sin1βsin3βsin5ββ―sin89β,
and let
Q=sin2βsin4βsin6ββ―sin88β.
Then
PQ=sin1βsin2βsin3ββ―sin89β.
But also
PQ=sin89βsin88βsin87ββ―sin1β.
Therefore
P2Q2β=sin1βsin89βsin2βsin88βsin3βsin87ββ―sin89βsin1β=sin1βcos1βsin2βcos2βsin3βcos3ββ―sin89βcos89ββ
Then
289P2Q2β=(2sin1βcos1β)(2sin2βcos2β)(2sin3βcos3β)β―(2sin89βcos89β)=sin2βsin4βsin6ββ―sin178β=(sin2βsin4βsin6ββ―sin88β)(sin92βsin94βsin96ββ―sin178β)=(sin2βsin4βsin6ββ―sin88β)(sin88βsin86βsin84ββ―sin2β)=Q2β
Thus 289P2Q2=Q2. Because Q2ξ =0, the requested product of cosecants equals P21β=289. Because 89 is prime, this representation with integers greater than 1 is unique. The requested sum is 2+89=91β.
The problems on this page are the property of the MAA's American Mathematics Competitions