MHB Can You Prove the Average of Trigonometric Numbers Equals Cot 1^o?

[anemone]

Prove that the average of the numbers $n\sin n^{\circ}$ (where $n=2,\,4,\,6,\, \cdots,\,180$) is $\cot 1^{\circ}$.

 

[kaliprasad]

Spoiler

because we are dealing with sin of even numbers

we have

$2n\, \sin\, 2n^0\, \sin\, 1^0 = n \cos (2n - 1)^0 - n \cos (2n+ 1)^0$

there are 90 terms

now adding from 1 to 90 we get the sum

=$ \cos\, 1^0 + \cos\, 3^0 + \cdots + \cos\, 179^0 - 90 \,cos\, 181^0$

as for all the terms except the 1st and last term

we have $- n\, \cos (2n+ 1)^0$ and $(n+1) \cos (2n+1)^0$ so $\cos (2n+1)^0$ has one occurence except $\cos\, 1^0$ (which comes ones because it is 1st term) and $\cos\, 179^0$ is with - 90

now as $\cos\, 1^0 + \cos\, 179^0 = 0$
$\cos\, 3^0 + \cos\, 177^0 = 0$

so on so the so the sum is

$- 90\, \cos\, 181^0$

which is same as

$90\, \cos\, 1^0$

so sum of given terms = $90\, \cos\, 1^0 / \sin\, 1^0 = 90\, \cot\, 1^0$ and as there are 90 terms average = $\cot\,1^0$