So I'm having problems with the double infinite potential well given by
$$V(x)= \left\{\begin{array}{ll} \infty & -\infty < x < -a-b \\ 0 & -a-b< x < -a \\ V_0 & -a < x < a \\ 0 & a < x < a + b \\ \infty & a+b < x < \infty \\ \end{array}\right.$$
I have to use the fact that the potential well is symmetric about x=0. I have solved the Schrödinger equation in all the regions and end up with
$$ψ1=A\sin(k_1x)+B\cos(k_1x)$$ $$ψ2=Ce^{k_2x}+De^{−k_2x}$$ $$ψ3=E\sin(k_1x)+F\cos(k_1x)$$ where $k_1=\sqrt{\frac{2mE}{\hbar^2}}$ and $k_2=\sqrt{\frac{2m(V_0-E)}{\hbar^2}}$. I can use the symmetry argument and get C=D which means cosh for even and sinh for odd wave functions. I know that I'm meant to get $A\sin(k_1(a+b−x))$ between the $-a-b$ and $-a$ and between $a+b$ and $a$ but I'm not sure how. If I know how to do that it may help me normalise the wave function which is key at the moment. Many thanks in advance!