# Relationship Between Trig Funtions and Bernoulli Numbers

1. Mar 26, 2014

### lwest513

1. The problem statement, all variables and given/known data
Prove the formula xcscx=2B(ix)-B(2ix)

2. Relevant equations
B(x)=x/((ex)-1)

sinx= (eix-e-ix)/2i
3. The attempt at a solution

I know that it makes sense to use the formula for B(x) with x=ix and x=2ix, and rewrite xcsc(x) as x/sin(x), plugging the above relevant equation in for sine. Manipulating these equations should bring about equality, but for some reason I've hit a road block where I can't come up with more algebraic manipulation to get to the equality.

I have that xcsc(x)= (2ix)/(eix-e-ix)
but after expanding 2B(ix)-B(2ix) using the B(x) formulas, I'm having a difficult time figuring out how to reduce it correctly and I'm too stuck in the same methods to see it a different way.
Right now I have (2ix(e2ix-eix))/(e3ix-eix-e2ix+1)

Any help would be greatly appreciated because I'm too caught up in a method that clearly isn't working!

EDIT:I figured it out but I'm not sure how to delete the post so I'll leave it. If anyone else is having the same problem, the key is to use the difference of squares on the B(2ix) term denominator.

Last edited: Mar 26, 2014
2. Mar 26, 2014

### lurflurf

hint factor two different ways

$$(e^{i x}-1)(e^{i x}+1)=(e^{2i x}-1)=e^{ix}(e^{i x}-e^{-i x})$$