$e^{-\beta \frac{1}{2}\hbar \omega} \frac{1}{1 - e^{\beta \hbar \omega}} = \frac{1}{2 \sinh \left( \frac{\beta \hbar \omega}{2}\right)}$

Question

\begin{equation} e^{-\beta \tfrac{1}{2}\hbar \omega} \dfrac{1}{1 - e^{\beta \hbar \omega}} = \dfrac{1}{2 \sinh \left( \frac{\beta \hbar \omega}{2}\right)} \end{equation} I need to know how this equality works?

Any help is really appreciated.

Answer 1

You have $$z= e^{-\beta h\omega /2} \frac 1 {1-e^{-\beta h\omega }}.$$ You just need to do the multiplication. Then

$$z= e^{-\beta h\omega /2} \frac 1 {1-e^{-\beta h\omega }} = \frac 1{e^{\beta h\omega /2}} \frac 1 {1-e^{-\beta h\omega }} = \frac 1 {e^{\beta h\omega/2 }-e^{-\beta h\omega/2 }}= \frac 1 {2 \sinh(\beta \omega h/2)},$$ because by definition $\sinh(x)= \frac 1 2 (e^x-e^{-x})$.