Maclurin Series. (Approximation) Given that $y=\ln \cos x$, show that the first non-zero terms of Maclurin's series for $y=-\frac{x^2}{2}-\frac{x^4}{12}$. Use this series to find the approximation in terms of $\pi$ for $\ln 2$.
My question is how to determine value of $x$ which is suitable?
 A: We have
\begin{align}
\ln(\cos(x)) & = \dfrac{\ln(\cos^2(x))}2 = \dfrac12 \cdot \ln(1-\sin^2(x)) = - \dfrac12 \sum_{k=1}^{\infty} \dfrac{\sin^{2k}(x)}k\\
& = -\dfrac12 \sin^2(x) - \dfrac14 \sin^4(x) - \cdots\\
& = -\dfrac12 \left(x-\dfrac{x^3}{3!} + \mathcal{O}(x^5)\right)^2-\dfrac14 \left(x + \mathcal{O}(x^3)\right)^4 + \mathcal{O}(x^6)\\
& = -\dfrac12\left(x^2 - \dfrac{x^4}6\right)  - \dfrac{x^4}4 + \mathcal{O}(x^6)\\
& = -\dfrac{x^2}2 - \dfrac{x^4}{12} + \mathcal{O}(x^6)
\end{align}
We need $\ln(2)$ to be expressed in terms of $\pi$, this means we need $\ln(\cos(x))$ to be some value related to $\ln(2)$. One such value is when $\cos(x) = 1/2$, which implies
$$\ln(1/2) = \ln(\cos(2n\pi \pm \pi/3)) = -\dfrac{(2n\pi \pm \pi/3)^2}2-\dfrac{(2n\pi \pm \pi/3)^4}{12} + \mathcal{O}\left((2n\pi\pm \pi/3)^6\right)$$
Since typically we are after good approximation, we would like to have the least possible error term, which in turn forces $n=0$, which gives us that
$$\ln(2) \approx \dfrac{\pi^2}{18} + \dfrac{\pi^4}{972}$$

Another way is to take $x=\pi/4$, which gives us that
$$-\dfrac12\ln(2) = \ln(1/\sqrt{2}) = \ln(\cos(\pi/4))) \approx - \dfrac{\pi^2}{32} - \dfrac{\pi^4}{3072}$$
This gives us that
$$\ln(2) \approx \dfrac{\pi^2}{16} + \dfrac{\pi^4}{1536} + \cdots$$
A: A suitable value is a value for which $\log\cos x=\log\frac{1}{2}$, for instance $x=\frac{\pi}{3}$.
That gives:
$$ \log 2 \approx \frac{\pi^2}{18}+\frac{\pi^4}{972}.$$
A better choice is $x=\frac{\pi}{4}$, that leads to:
$$ \log 2 \approx \frac{\pi^2}{16}+\frac{\pi^4}{1536}.$$
A: $$\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$$
$$\ln\cos(\pi/3)=\ln(1/2)=\ln(1)-\ln(2)=-\ln(2)$$
