2D standing wave When we have 1D standing waves, we can write them as the sum of two propagating wave in opposite directions that give the formula $\sin(kx)\cos(wt)$.
When I try to do this for 2D waves (I mean 2D by the fact there are $k_x$ and $k_y$) I don't have the right expression :
$$\exp\left(i\left(k_x x + k_y y - wt\right)\right)+\exp\left(i\left(k_x x + k_y y + wt\right)\right)=\exp\left(i\left(k_x x + k_y y\right)\right)2\cos(wt)$$
If I take the imaginary part I will have $\sin(k_x x +k_y y)$ and not $\sin(k_x x)\sin(k_y y)$.
Am I wrong somewhere or we can't say that 2D stationary waves are the sum of two propagating waves in opposite directions ?
 A: So, you summed two waves and got $[\sin(k_xx)\cos(k_yy)+\cos(k_xx)\sin(k_yy)]\cos(\omega t)$. Now, add two more with wavevector $(k_x,-k_y)$ and you will obtain $2\sin(k_xx)\cos(k_yy)\cos(\omega t)$, the standing wave you're looking for.
However, the function $\sin(k_xx+k_yy)\cos(\omega t)$ already is a standing wave, since the spatial and the temporal oscillations are independent: $f(x,y)g(t)$. Here nodes and anti-nodes consist of lines perpendicular to $\vec{k}$.
Experimentally, a standing wave results from the superposition of incident and reflected waves. In one dimension you need one wall (one reflection). In two dimensions you generally need two non-parallel walls; the second wall will reflect the incident wave and the wave reflected on the first wall, hence the need of four travelling waves to make a standing one.
But, if you place a single wall perpendicularly to $\vec{k}$, you obtain (in 2D- or 3D-space) a one-dimensional standing wave, it just has a peculiar orientation. Experimentally this can be approximately obtained by placing a loudspeaker facing a wall.
