I am trying to approximate a function $f(r) = \sqrt{r^2 +a^2 +b^2 -2ra\cos \theta \sin \phi -2rb \cos \phi}$.
I want to use $r\times\sqrt{1+x} \approx r(1 +x/2 -x^2/8)$ to approxiimate this.
Here is what I did: I assumed $x = \frac{a^2+b^2-2ar \cos \theta \sin \phi -2br\cos \phi}{r^2}. $
Hence, $\sqrt{r^2 +a^2 +b^2 -2ra\cos \theta \sin \phi -2rb \cos \phi} = r\times \sqrt{1+x}$.
Therefore, $\sqrt{r^2 +a^2 +b^2 -2ra\cos \theta \sin \phi -2rb \cos \phi} =\\ r \times\left(1+ \frac{(a^2+b^2-2ar \cos \theta \sin \phi -2br\cos \phi)}{2r^2}- \frac{a^4}{8r^4} -\frac{b^4}{8r^4}-\frac{a^2\cos^2\theta \sin^2 \phi }{2r^2}-\frac{b^2\cos^2 \phi}{2r^2}-\frac{a^2b^2}{4r^4} +\frac{a^3 \cos\theta \sin\phi }{2r^3}+\frac{a^2b\cos\phi}{2r^3}+\frac{ab^2\cos \theta \sin \phi}{2r^3}+\frac{b^3cos\phi}{2r^3}-\frac{ab\cos\theta\sin\phi\cos\phi}{r^2}\right)$. Now, this can be rewritten as
$f(r) = r - a \cos\theta \sin \phi -b \cos \phi +\frac{a^2}{2r}[1-\cos^2\theta \sin^2 \phi] + \frac{b^2}{2r^2}[\sin^2\phi] -\frac{ab \cos \theta \sin \phi \cos \phi}{r}+ \frac{1}{2r^2}[a^3\cos\theta \sin\phi + b^3 \cos \phi +a^2b\cos\phi +ab^2\cos\theta\sin\phi]-\frac{1}{8r^3}[a^2+b^2]^2.$
Can somone help me verify If I did any mistakes in this?