Here, Wikipedia has the derivation for the equation:
$$\cos(ka)=\cos(\alpha a) + P \frac{\sin(\alpha a)}{\alpha a}$$
I didn't understand one of the steps:
$\beta^2-\alpha^2$ should be equal to $\frac{2m(V_o-2|E|)}{\hbar^2}$ and not $\frac{2m(V_o)}{\hbar^2}$. I don't understand how they managed to come up with $P=\frac{mV_o ba}{\hbar^2}$. Am I missing something?
$$\cos(ka)=\cos(\beta b)\cosh[\alpha(a-b)]+\frac{\beta^2-\alpha^2}{2\alpha\beta}\sin(\beta b)\sinh[\alpha(a-b)]$$ $$\cos(ka)=\cos(\beta b)\cosh[\alpha(a-b)]+\frac{\beta^2(1-\frac{\alpha^2}{\beta^2})b}{2\alpha}\frac{\sin(\beta b)}{\beta b}\sinh[\alpha(a-b)]$$
Approximate for $b\to 0\quad\implies\quad\beta\to\infty\quad;\quad \beta b\to 0\quad$ (See the page in Wikipedia) :
$\frac{\sin(\beta b)}{\beta b}\to 1 \quad ;\quad 1-\frac{\alpha^2}{\beta^2}\to 1$
$$\cos(ka)\simeq\cos(\beta b)\cosh[\alpha(a-b)]+\frac{\beta^2 b}{2\alpha}\sinh[\alpha(a-b)]$$
$\cosh[\alpha(a-b)]\to\cosh(\alpha a) \quad ;\quad \sinh[\alpha(a-b)]\to\sinh(\alpha a)$ $$\cos(ka)\simeq\cos(\beta b)\cosh(\alpha a)+\frac{\beta^2 b}{2\alpha}\sinh(\alpha a)$$ $\beta b\to 0 \implies \cos(\beta b)\to 1\quad$ See the page in Wikipedia. $$\cos(ka)\simeq\cosh(\alpha a)+\frac{\beta^2 b}{2\alpha}\sinh(\alpha a)$$ $$\cos(ka)\simeq\cosh(\alpha a)+\frac{\beta^2 ba}{2}\frac{\sinh(\alpha a)}{\alpha a} = \cosh(\alpha a)+P\frac{\sinh(\alpha a)}{\alpha a}$$ $$P=\frac{\beta^2 ba}{2}$$ $\beta^2=\frac{2m(V_0-|E|)}{\hbar^2}= \frac{2mV_0(1-\frac{|E|}{V_0})}{\hbar^2} \to \frac{2mV_0}{\hbar^2}$ $$P\simeq\frac{2mV_0}{\hbar^2}\frac{ ba}{2} = \frac{mV_0 ab}{\hbar^2}$$