Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.
given that \(\displaystyle\sum_{k=1}^{35}sin5k=tan\frac{m}{n}\) where angles are measured in degrees, m and n are relatively prime positive integer that satisfy \(\frac{m}{n} \lt 90\), find m+n.
Angles
Triangles
Side Length
Answer: is 177.
AIME I, 2009, Question 5
Plane Trigonometry by Loney
s=\(\displaystyle\sum_{k=1}^{35}sin5k\)
s(sin5)=\(\displaystyle\sum_{k=1}^{35}sin5ksin5=\displaystyle\sum_{k=1}^{35}(0.5)[cos(5k-5)-cos(5k+5)]\)=\(\frac{1+cos5}{sin5}\)
\(=\frac{1-cos(175)}{sin175}\)=\(tan\frac{175}{2}\) then m+n=175+2=177.
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