Problem:
For π≤θ<2π, let
P=21cosθ−41sin2θ−81cos3θ+161sin4θ+321cos5θ−641sin6θ−1281cos7θ+…
and
Q=1−21sinθ−41cos2θ+81sin3θ+161cos4θ−321sin5θ−641cos6θ+1281sin7θ+…
so that QP=722. Then sinθ=−nm where m and n are relatively prime positive integers. Find m+n.
Solution:
Notice that
Q+iP=1+21(−sinθ+icosθ)+41(−cos2θ−isin2θ)+81(sin3θ−icos3θ)+⋯=1+21ieiθ+41i2e2iθ+81i3e3iθ+⋯=2−ieiθ2=5+4sinθ2(2+sinθ+icosθ)
Thus P=5+4sinθ2cosθ and Q=5+4sinθ4+2sinθ. So
722=QP=2+sinθcosθ32+32sinθ+8sin2θ=49−49sin2θ57sin2θ+32sinθ−17=(3sinθ−1)(19sinθ+17)=0
and
sinθ=−1917 or 31
The requested sum for the negative solution is 17+19=36.
The problems on this page are the property of the MAA's American Mathematics Competitions