# Maclaurin expansion problem

I am having problem with a question regarding Maclaurin expansion.

a) Find the term up to $x^4$ in the Maclaurin expansion of $f(x)=\ln(\cos(x))$.

This part I was able to anwser part a. The Maclaurin expansion until $x^4$ is: $f(x)=-x^2/2-x^4/12$

b) Use this series to find an approximation in terms of $\pi$ for $\ln(2)$.

I was not able to answer part b. If someone know how to make it I would appreciate it.

Thanks

Hint: Use the fact that $\ln(a^b)=b\cdot\ln(a)$. Therefore, $\ln(2)=-\ln\bigl(\frac{1}{2}\bigr)$. Then, when does $\cos(x)=\frac{1}{2}$?
a) From $$\begin{eqnarray*}\log\cos x &=& \log\prod_{n\geq 0}\left(1-\frac{4x^2}{\pi^2(2n+1)^2}\right)\\&=&-\sum_{n\geq 0}-\log\left(1-\frac{4x^2}{\pi^2(2n+1)^2}\right)\\&=&-\sum_{n\geq 0}\sum_{m\geq 1}\frac{4^m x^{2m}}{m\pi^{2m}(2n+1)^{2m}}\\&=&-\sum_{m\geq 1}\frac{(4^m-1)\zeta(2m)\,x^{2m}}{m\pi^{2m}}\end{eqnarray*}$$ we have all the coefficients of the Taylor series at the origin,
b) and by evaluating both sides of $$\log\cos(x) \approx -\left(\frac{x^2}{2}+\frac{x^4}{12}+\frac{x^6}{45}\right)$$ at $x=\frac{\pi}{4}$ we get $$\log(2)\approx \frac{\pi ^2}{16}+\frac{\pi ^4}{1536}+\frac{\pi ^6}{92160}.$$
Actually the (Padé) approximation $\log\cos x \approx -\frac{x^2}{2 \left(1-\frac{x^2}{6}\right)}$ has a comparable accuracy and leads to $$\log(2) \approx \frac{6\pi^2}{96-\pi^2}.$$