299/ Potw

For a triangle ABC with the sides of a,b,c such that a + c = 2b prove that $\cot(\frac{A}{2}) + \cot(\frac{C}{2}) = 2 \cot(\frac{B}{2})$

Solution:
We have using law of cot for a triangle using standard notation s ,and area of triangle $\triangle$

$\cot(\frac{A}{2}) = \dfrac{s(s-a)}{\triangle}\cdots(1)$

$\cot(\frac{B}{2}) = \dfrac{s(s-b)}{\triangle}\cdots(2)$

$\cot(\frac{C}{2}) = \dfrac{s(s-c)}{\triangle}\cdots(3)$

add (1) and (3) to get

 $\cot(\frac{A}{2}) + \cot(\frac{C}{2}) = \dfrac{s(s-a)}{\triangle} + \dfrac{s(s-c)}{\triangle}$
$= \dfrac{s(s-a)+s(s-c)}{\triangle}$
$= \dfrac{s(s-a+s-c)}{\triangle}$
$= \dfrac{s(2s-(a+c))}{\triangle}$
$= \dfrac{s(2s-2b)}{\triangle}$ (from given conditon)
$= 2\dfrac{s(s-b)}{\triangle}$
$=2\cot(\frac{B}{2})$ (using (2))

proved 

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