Finding potential using spherical harmonics I have been trying to solve the following question: 
The potential on the surface of a sphere is given by
$\mathbf {V = V_{0} \sin^2\theta \sin2\phi,\;}$
find the potential outside the sphere
I am trying to solve it by separation of variable in spherical coordinates by using  the following formula for potential outside the sphere,
$$V=\sum_{l=0}^\infty\frac{B_{lm}}{r^{l+1}} {Y_l}^m (\theta,\phi)$$
Now the potential on the surface of the sphere is given, so we can use that for r=R as,
$$\tag{1}V_{0}\sin^2\theta\sin 2\phi=\sum_{l=0}^\infty\frac{B_{lm}}{R^{l+1}} {Y_l}^m (\theta,\phi)$$
Next for the value of $B_l$ I multiply both side with ${Y^*_l}^m$ and integrate. RHS becomes $\frac{B_{lm}}{r^{l+1}}$ while LHS becomes interesting. I note that $\sin^2\theta$ $\sin 2\phi$ can be converted into $Y_2^2$ with some Constant factor. $Y_2^2$ is given as follows: $$ Y_2^2= A \sin^2\theta\ e^{im\phi}$$ 
So my problem is, can I some how convert this into $Y_2^2$ so that it simply gives me the left hand side of equation ${(1)}?\;$ I see that $sin2\phi$ is the imaginary part of $e^{im\phi}$ with $m=2$. Please guide me through this.
 A: You already noticed $\sin 2\phi$ is the imaginary part of $e^{2i\phi}$.
Another way to say this is
$$\sin 2\phi=\frac{i}{2}\left(-e^{2i\phi}+e^{-2i\phi}\right)$$
From the table of spherical harmonics you have:
$$Y_2^{+2}(\theta,\phi)=\frac{1}{4}\sqrt{\frac{15}{2\pi}}\sin^2\theta \ e^{2i\phi}$$
$$Y_2^{-2}(\theta,\phi)=\frac{1}{4}\sqrt{\frac{15}{2\pi}}\sin^2\theta \ e^{-2i\phi}$$
Putting this together you find (even without doing any integral):
$$\sin^2\theta\ \sin 2 \phi=
2i\sqrt{\frac{2\pi}{15}} \left(-Y_2^{+2}(\theta,\phi) + Y_2^{-2}(\theta,\phi)\right)$$
You see, one spherical harmonic was not enough.
You needed two of them.
A: What if you assume that the initial potential has its imaginary component, just for the sake of the argument Vo $\sin^2\theta$ $\sin 2\phi$ = Re{$U_0Y_2^2$} where $U_0$ is some constant you get from the definition of your potential and $Y_2^2$? Then you indeed can prove that the potential outside the sphere is the same $Y_2^2$ with some constant factor. The imaginary part has no physical meaning then, you use only the real one since all equations must hold if you apply Re{} or Im{} to them.
My guess you will come to something like C$\cdot$Vo $\sin^2\theta$ $\sin 2\phi / r^3$ 
Spherical harmonics are orthonormal. If you have a spherical harmonic on one side, you have to have the same one on the other, no more and no less. 
