= Solution
Reflecting the solution across $\theta=\pi\alpha$ preserves its <Dirichlet boundary conditions> and outer boundary data, so uniqueness makes the solution reflection-symmetric. Since
$$
\sin\frac{n(2\pi\alpha-\theta)}{2\alpha}=(-1)^{n+1}\sin\frac{n\theta}{2\alpha},
$$
only odd angular <Fourier modes> survive. Put $\rho=(r/R)^{1/(2\alpha)}$ and $\psi=\theta/(2\alpha)$. On the symmetry axis $\psi=\pi/2$, comparison of the convergent <power series> gives
$$
A_{2j+1}R^{(2j+1)/(2\alpha)}(-1)^j=\frac1{(2j+1)!}.
$$
Thus
$$
\boxed{\Phi(r,\theta)=\sum_{j=0}^{\infty}\frac{(-1)^j\rho^{2j+1}}{(2j+1)!}\sin\bigl((2j+1)\psi\bigr).}
$$
This is the <imaginary part> of $\sin(\rho e^{i\psi})$, so a useful closed form is
$$
\boxed{\Phi(r,\theta)=\cos\bigl(\rho\cos\psi\bigr)\sinh\bigl(\rho\sin\psi\bigr).}
$$
At either straight edge $\sin\psi=0$, while on the symmetry axis the formula is $\sinh\rho$. Every term is <harmonic> in the sector, and the factorial denominator gives convergence up to $r=R$, verifying the full solution.
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