Problem:
Given that ∑k=135sin5k=tannm, where angles are measured in degrees, and m and n are relatively prime positive integers that satisfy nm<90, find m+n.
Solution:
k=1∑35sin5k=sin51k=1∑35sin5sin5k=sin51k=1∑352cos(5k−5)−cos(5k+5)=sin51⋅2cos0+cos5−cos175−cos180=sin1751−cos175=tan2175
so m+n=177.
OR
Let cis t=cost+isint. Because cis 0=1, the given series is the imaginary part of the complex series
cis 0+cis 5+cis 10+…+cis 175=k=0∑35(cis 5)k,
by the theorem of DeMoivre. This is a geometric series, whose sum is
1−cis 5cis 0−cis 180=1−cis 52=1−cos5−isin52=(1−cos5)2+(sin5)22(1−cos5+isin5)
The imaginary part of the sum is
1−cos5sin5=1+cos175sin175=tan2175.
Thus m+n=177.
The problems on this page are the property of the MAA's American Mathematics Competitions