Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.
Series and sum – AIME I, 1999
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.
- is 107
- is 177
- is 840
- cannot be determined from the given information
Key Concepts
Angles
Triangles
Side Length
Check the Answer
Answer: is 177.
AIME I, 2009, Question 5
Plane Trigonometry by Loney
Try with Hints
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.
Other useful links
- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA