The Helmholtz equation $(\nabla^2\phi=-k^2\phi)$ has been solved using trigonometric series for a rectangular geometry $a>x>0$ and $b>y>0$ subject to some boundary condition. To be specific, the normal derivative at the boundary is zero, except at $d>x>c$ and $y=0$ where the tangential component of $\nabla \times \phi$ has a constant value. Particulars of the boundary conditions are not important.
Two equivalent solutions are found.
$$\phi=\sum_{m,n}A_{mn}\cos(\frac{m \pi}{a}x)\cos(\frac{n \pi}{b}y)=\sum_{m}B_{mn}\cos(\frac{m \pi}{a}x)\cos\bigg(\sqrt{k^2-(\frac{m \pi}{a})^2}(y-b) \bigg)$$
Question 1: How on earth could the first series satisfy the Helmholtz equation? Taking the Laplacian of the series is equivalent to multiplying each term by $-k_{mn}^2=-(\frac{m\pi}{a})^2-(\frac{n\pi}{b})^2$. In order to solve the Helmholtz equation, this modified series should equal the original series times $-k^2$.
$$\sum_{m,n}k_{mn}^2A_{mn}\cos(\frac{m \pi}{a}x)\cos(\frac{n \pi}{b}y)=k^2\sum_{m,n}A_{mn}\cos(\frac{m \pi}{a}x)\cos(\frac{n \pi}{b}y)$$
I can't wrap my head around how this is possible. I thought these cosine functions were orthogonal and in order for two series to be equal, they had to match term by term. Clearly, they don't. Perhaps this is one of those mysterious scenarios where one isn't allowed to swap the order of summation and integral/derivative operators...
Question 2: How can the second series guarantee normal derivative vanishes at the boundary? In the first series, cosines that perfectly fit are used, which guaranteed the normal derivative will be 0 at the boundary. Second series has no such property. It adds up a bunch of cosines (and hyperbolic cosines) that do not perfectly fit in the region $b>y>0$ and we just hope that it'll satisfy the boundary conditions. The second series has the advantage of being rigged such that it will satisfy the Helmholtz equation term by term.
Funny thing is, I can see myself solving the problem through either method, but when I think about it deeply, it starts to feel like it shouldn't work. I tried to test the series numerically and they both seem to be equal and satisfy the Helmholtz equation. Clearly I have some misconceptions.