PROBLEM
Let us consider the following Dirichlet Problem for the Laplace Equation:
$$ r^2 \partial_r^2(u) + r \partial_r(u) + \partial_{\theta}^2(u) = 0, \hspace{5mm} (r,\theta) \in ]0,1[ \times ]0, \pi/2[ $$ $$ u(1,\theta) = \sin(2\theta)^2, \hspace{5mm} \theta \in [0, \pi/2] $$ $$ u(r,0) = u(r,\pi/2) = 0, \hspace{5mm} r \in [0,1] $$
MY ATTEMPT
We know that the solution to this problem, if it exists, is unique. Therefore we only need to find one function that satisfies both the boundary conditions and the equation to solve it. My idea was then to develop $u$ into a Fourier Series of sines and cosines with respect to the variable $\theta$ with fixed $r$, just as I would do if the problem was given on the whole unitary disk. However, in order for the second boundary condition to be satisfied, I decided to only keep the terms of the form $\sin(2n\theta), n\in \Bbb Z_+$:
$$ u(r,\theta) = \sum_{n=1}^{+\infty}b_n r^{2n} \sin(2n\theta) $$
Now, I only need to find the coordinates $b_n$ by using the first boundary conditions. To this purpose I developed $\sin(x)^2$ in a Fourier Series of sines in $[0,\pi]$ (this should be possible since I can extend the function to be odd in $[0,2\pi]$ with integrable derivative):
$$ \sin(x)^2 = \sum_{n=0}^{+\infty} \frac{1}{\pi} \left( \frac{1}{(2n+1)+2} - \frac{2}{(2n+1)} + \frac{1}{(2n+1)-2} \right) \sin((2n+1)x), \hspace{5mm} x \in [0,\pi] $$
(only the even terms survived when I solved the integrals for the coordinates). Evaluating the equation for $x=2\theta$ we obtain an expression for $\sin(2\theta)^2$ (which I guess is just its Fourier Series of sines in $[0,\pi/2]$) which can be used to determine the $b_n$'s by comparison:
$$ b_{2(2n+1)} = \frac{1}{\pi} \left( \frac{1}{(2n+1)+2} - \frac{2}{(2n+1)} + \frac{1}{(2n+1)-2} \right), \hspace{5mm} n \in \Bbb N $$ $$ b_n = 0 \hspace{5mm} \text{in all the other cases} $$
MY QUESTION
I would like to know if the procedure and the solution are correct and wether my argument can be improved.