what is the taylor expansion of $\ln(\sin x/x)$ what is the Taylor expansion of the function
$$f(x) = \ln \frac{\sin x}x$$
around  the point $x=0$? Ignore powers of $x$ which are greater than $6$.
Here is my method:
$$\ln(1+x)=x-\frac{x^2}2 + \frac{x^3}3 -\frac{x^4}4,$$
so we should get the function
$$g(x)=\frac{\sin x}x$$
in the form of $1 + x$ to get the answer … or is there another way to consider the Taylor expansion of $\ln (u)$ by considering $f$ to be the answer of $$\frac{\sin x}x\cdots$$
Which way is the best?? What is the final answer???  
 A: $$\begin{align}\log(\frac{\sin x}{x}) 
= & \log\prod_{n=1}^{\infty}\left( 1 - \frac{x^2}{n^2\pi^2}\right)
= \sum_{n=1}^{\infty}\log\left(1 - \frac{x^2}{n^2\pi^2}\right)\\
\stackrel{\color{blue}{^{[1]}}}{=} & -\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{x}{n\pi}\right)^{2k}
\stackrel{\color{blue}{^{[2]}}}{=} -\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{k}\left(\frac{x}{n\pi}\right)^{2k}\\
\stackrel{\color{blue}{^{[3]}}}{=} & -\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k\pi^{2k}} x^{2k}
= -\frac{1}{6}x^2 -\frac{1}{180}x^4 - \frac{1}{2835} x^6 - \frac{1}{37800} x^8 - \cdots
\end{align}$$
Notes
$\color{blue}{[1]}$ the power series converges when $|x| < \pi$.
$\color{blue}{[2]}$ legal to change order of summation because all terms have same sign.
$\color{blue}{[3]}$ $\zeta(z) = \sum_{n=1}^{\infty} n^{-z}$ is the Riemann zeta function. When $z$ is an even integer, it can be expressed in terms of Bernoulli number $B_{2k}$:
$$\zeta(2k) = \frac{(-1)^{k+1} B_{2k} (2\pi)^{2k}}{2(2k)!}$$
Look at the wiki pages for more relations.
A: As you suggest, it will be useful to know the expansion
$$\ln(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots\tag{1}$$
(at least for $|t|\lt 1$).
We also need the power series expansion for $\sin x$, which is
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots.$$
So the function which is $1$ at $x=0$ and $\frac{\sin x}{x}$ elsewhere has power series expansion 
$$1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots.\tag{2}$$
Now substitute (2) for $1+t$ in (1). Thus $t=-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots$.
Make  sure not to compute too far. So throw away all terms in powers of $x$ greater than $6$.
Because the expression (2) has no $x$ term, you will only need to compute to the $\frac{t^3}{3}$ term in (1). So the calculation is less time-consuming than it seems. 
Getting a general term for the power series expansion looks, however, very difficult. Luckily you were not asked to find many terms. You will end up with something of the shape $ax^2+bx^4+cx^6$.
A: Your instincts are correct.  Write
$$\frac{\sin{x}}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots$$
We wish to maintain the series to $O(x^6)$.  Then
$$\begin{align}\ln{\left ( \frac{\sin{x}}{x}\right)} &=  \left (\frac{\sin{x}}{x} - 1\right ) - \frac12 \left (\frac{\sin{x}}{x} - 1\right )^2 + \frac13 \left (\frac{\sin{x}}{x} - 1\right )^3-\cdots \\ &= \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right ) - \frac12 \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right )^2+\\ &\frac13 \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right )^3-\cdots \\ &= - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots - \frac{x^4}{72} \left (1 - \frac{x^2}{10}+\cdots \right ) - \frac{x^6}{648}+\cdots \\ &= - \frac{x^2}{6} - \frac{x^4}{180}-\frac{x^6}{2835}+\cdots \end{align}$$
Note that
$$-\frac{1}{180} = \frac{1}{120} - \frac{1}{72}$$
$$-\frac{1}{2835} = -\frac{1}{5040}+\frac{1}{720}-\frac{1}{648}$$
