Laplace Transform: $\frac{1}{s^2} \left( 1 - e^{-s t_0} \right)^2 = \frac{4}{s^2} e^{-st_0} \sinh^2 \left( \frac{1}{2}st_0 \right)$

Question

I am trying to verify my textbook's solution for the Laplace transform of

$$F(t)= \begin{cases} t & 0 \le t < t_0\\ 2t_0 - t & t_0 \le t \leq 2t_0\\ 0 & t > 2t_0 \end{cases}$$

$$\begin{align} \mathcal{L} \{F(t)\} &= \int_0^{t_0} te^{-st} \ dt + \int_{t_0}^{2t_0} (2t_0 - t)e^{-st} \ dt \\ &= \dfrac{1}{s^2} \left( 1 - e^{-s t_0} \right)^2 \\ &= \dfrac{4}{s^2} e^{-st_0} \sinh^2 \left( \dfrac{1}{2}st_0 \right) \end{align}$$

I was able to verify everything up to $\dfrac{1}{s^2} \left( 1 - e^{-s t_0} \right)^2$, but I don't see how $\dfrac{1}{s^2} \left( 1 - e^{-s t_0} \right)^2 = \dfrac{4}{s^2} e^{-st_0} \sinh^2 \left( \dfrac{1}{2}st_0 \right)$?

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

$$(1-e^{-x})^2 = 4\bigg(\frac{1-e^{-x}}{2}\bigg)^2 = 4e^{-x}\bigg(\frac{e^{x/2}-e^{-x/2}}{2}\bigg)^2 = 4e^{-x}\sinh^2\bigg(\frac{x}{2}\bigg)$$

Here $x = st_0$