Wavefunction in a delta potential well

[1v1Dota2RightMeow]

Homework Statement

Using the equations given, show that the wave function for a particle in the periodic delta function potential can be written in the form

$\psi (x) = C[\sin(kx) + e^{-iKa}\sin k(a-x)], \quad 0 \leq x \leq a$

Homework Equations

Given equations:

$\psi (x) =A\sin(kx) + B\cos(kx), \quad 0<x<a$
$A\sin(ka) = [e^{iKa} - \cos(ka)]B$

Note that $k$ and $K$ are different constants.

The Attempt at a Solution

I tried a bunch of stuff already but I can't seem to get to the answer.

Attempt 1. I evaluated $\psi$ at $0$ and at $a$ for both the final equation and the general equation and tried to see if I could come to some conclusion based on equating these, but no luck there.

Attempt 2. I tried working backwards and seeing if I could use the sine identity $\sin(a-b) = \sin(a)\cos(b)-\cos(a)\sin(b)$ but it only seems to make things more complicated.

Could someone just give me a hint?

 

[vela]

Try multiplying $\psi(x)$ by $\frac{\sin ka}{\sin ka}$.

 

[1v1Dota2RightMeow]

Try multiplying $\psi(x)$ by $\frac{\sin ka}{\sin ka}$.

I see 3 ways to do something with what you've suggested. Here is one attempt:

$\psi(x) = \frac{Asin(kx)sin(ka)+Bcos(kx)sin(ka)}{sin(ka)}$
$=Asin(kx) + \frac{B[(1/2)(sin(ka+kx)+sin(ka-kx))]}{sin(ka)}$
$=Asin(kx)+\frac{A}{2(e^{iKa}-cos(ka))}[sin(ka+kx)+sin(ka-kx)]$
$=Asin(kx)+\frac{e^{-iKa}A}{2(1-e^{-iKa}cos(ka))}[sin(ka+kx)+sin(ka-kx)]$

This would almost be great if it weren't for that $sin(ka+kx)$ term. I don't know what to do with it.

 

[vela]

There's a reason I said to multiply $\psi$ by $\frac{\sin ka}{\sin ka}$ rather than just the last term. See what you can do with the first term.

 

[1v1Dota2RightMeow]

There's a reason I said to multiply $\psi$ by $\frac{\sin ka}{\sin ka}$ rather than just the last term. See what you can do with the first term.

I expanded it out to this, but nothing cancels nicely.

$\psi (x) = \frac{A(sin(kx)sin(ka)-cos(kx)cos(ka))}{sin(ka)}+\frac{B((1/2)(sin(kx)cos(ka)+cos(kx)sin(ka)-sin(kx-ka)))}{sin(ka)}$

Should I have gone a different route?