<s>[INST] Taylor expansion of $ \arccos(\frac{1}{\sqrt{2}}+x)$, $ x\rightarrow0$ What is the method to calculate the Taylor expansion of $ \arccos(\frac{1}{\sqrt{2}}+x)$, $ x\rightarrow0$ ?
 [/INST] The formula for the cosine of a difference yields
$$
\begin{align}
\cos(\pi/4-y)
&= \frac{1}{\sqrt{2}}\cos(y)+\frac{1}{\sqrt{2}}\sin(y)\\
&=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}(\sin(y)+\cos(y)-1)\\
&=\frac{1}{\sqrt{2}}+x\tag{1}
\end{align}
$$
Noting that $x=\frac{1}{\sqrt{2}}(\sin(y)+\cos(y)-1)$, it is easy to show that 
$$
2\sqrt{2}x+2x^2=\sin(2y)\tag{2}
$$
Now the series for $\sin^{-1}(x)$ can be gotten by integrating the series for $\dfrac{1}{\sqrt{1-x^2}}$. Using the binomial theorem, we get
$$
(1-x^2)^{-\frac{1}{2}}=\sum_{k=0}^\infty\binom{2k}{k}\frac{x^{2k}}{4^k}\tag{3}
$$
Integrating $(3)$, we get
$$
\sin^{-1}(x)=\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}\frac{x^{2k+1}}{4^k}\tag{4}
$$
Combining $(1)$, $(2)$, and $(4)$, we get that
$$
\begin{align}
\cos^{-1}\left(\frac{1}{\sqrt{2}}+x\right)
&=\frac{\pi}{4}-y\\
&=\frac{\pi}{4}-\frac{1}{2}\sin^{-1}(2\sqrt{2}x+2x^2)\\
&=\frac{\pi}{4}-\frac{1}{2}\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}\frac{(2\sqrt{2}x+2x^2)^{2k+1}}{4^k}\\
&=\frac{\pi}{4}-\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}(\sqrt{2}x+x^2)^{2k+1}\tag{5}
\end{align}
$$
To get $2n$ terms of the Taylor series for $\cos^{-1}\left(\frac{1}{\sqrt{2}}+x\right)$, you only need $n$ terms of $(5)$.
Afterthought:
A nicer series, that doesn't involve all the $\sqrt{2}$s would be
$$
\cos^{-1}\left(\frac{1+x}{\sqrt{2}}\right)=\frac{\pi}{4}-\sum_{k=0}^\infty\frac{1}{2k+1}\binom{2k}{k}(x+\tfrac{1}{2}x^2)^{2k+1}
$$
</s>