298/ Secondary school/ High school students

Problem:

Given that $ \sum_{k=1}^{35}\sin\,5k =\tan(\frac{a}{b}) $ where angles are measured in degrees and a and b are relatively prime positive integers that satisfy $\frac{a}{b} < 90$ evaluate a + b


Solution

We have by sum of sin kx as below

$$ \sum_{k=1}^{n}\sin\,kx = \frac{\sin\frac{nx}{2}\sin\frac{(n+1)x}{2}}{\sin \frac{x}{2}}$$

put n = 35 and $ x = 5^\circ$


to get $$ \sum_{k=1}^{35}\sin\,5k^\circ = \frac{\sin\frac{175}{2}^\circ\sin\,90^\circ}{\sin \frac{5}{2}^\circ}$$
$$ = \frac{\sin\frac{175}{2}^\circ}{\cos \frac{175}{2}^\circ}=\tan  \frac{175}{2}^\circ$$

from given condition $a=175,b=2,\frac{a}{b} = 87.5 <90$ so a+b = 177