$F(x)=A\sin(kx)+B\cos(kx)$
$G(x)=C\sin(k(a-x))+D\cos(k(a-x))$
Applying physical boundary conditions gives us
$B=D=0$. I have another condition
$F(b)=G(b)$. To satisfy this condition I should have
$A\sin(kb)=C\sin(k(a-b))$
I need the values of A and C. Can I take the value $A=\sin(k(a-b))$ and $C=\sin(kb)$