Suppose we have a semidisk, with the origin in the center of it. I have to solve the Laplace equation
$$\nabla^2u(\rho,\phi)=0 \tag{1}$$
with the following boundary conditions
$$u(r,0)=u(r,\pi)=T_0\tag{2}$$ $$u(R,\phi)=T_1\tag{3}$$ The general solution to Laplace's equation that is regular at $\rho=0$ can be written as
$$u(\rho,\phi)=a_0+\sum_{n=1}^{\infty}\rho^n[a_n\cos(n\phi)+b_n\sin(n\phi)]. \tag{4}$$
Applying $u(r,0)=T_0$ we get
$$u(\rho,0)=a_0+\sum_{n=1}^{\infty}\rho^n a_n = T_0 \tag{4} $$
The only way for that to be true for all $\rho\in[0,R)$ is that $a_n=0$ for $n\geq 1$. That way, $(4)$ reduces to
$$u(\rho,\phi)=T_0+\sum_{n=1}^{\infty}\rho^n b_n\sin(n\phi). \tag{5}$$
I'm stuck at this point. I don't know how to apply the condition $(3)$ to $(5)$ in order to solve for $b_n$. Any suggestion will be really appreciated.
Edit: As whpowell96 mentioned, if I apply the condition I get
$$u(R,\phi)=T_0+\sum_{n=1}^{\infty}R^n b_n\sin(n\phi). \tag{5}$$
In other words
$$\sum_{n=1}^{\infty}R^n b_n\sin(n\phi) = T_1-T_0 \tag{6}$$
If now we compute the Fourier transform of $T_1-T_0$:
$$T_1-T_0 = A_0 + \sum_{n=1}^{\infty}\left( A_n \cos(n\phi) +B_n\sin(n\phi)\right) \tag{7}$$
We can identify terms and say that
$$R^n b_n = B_n = \frac{1}{\pi}\int_0^{2\pi}(T_1-T_0)\sin(n\phi) d\phi \tag{8}$$
But that's only possible if $b_n=0$, and that implies that $u(\rho,\phi)=T_0$, which is wrong. What I' missing here?