By creating a Maclaurin series up to an including $x^4$ for $\ln(\cos x)$ shows that $\ln2 \approx \frac{\pi^2}{16}\left( 1+\frac{\pi^2}{96}\right)$

By creating a Maclaurin series up to an including $$x^4$$ for $$\ln(\cos x)$$ shows that $$\ln2 \approx \frac{\pi^2}{16}\left( 1+\frac{\pi^2}{96}\right)$$

So creating a Maclaurin series using the general formula using derivatives when $$x=0$$, is got $$\ln(\cos x)\approx -\frac12x^2 - \frac1{12}x^4$$

I have seen that this question has already been asked in a different post, however they give in the question that $$x=\frac{\pi}{4}$$ with no reason for that choice. I can see that using that value works, but is there a reason for this choice particularly as it doesn't seem a logical first choice to me or is there another way of doing this with complex numbers.

• Because $\cos(\pi/4)=\frac 1{\sqrt2}$ – Claude Leibovici Apr 18 at 9:49

We have $$\cos x \ne 2$$ for all real $$x$$, hence we are looking for $$s,x \in \mathbb R$$ such that
$$s \cdot \ln ( \cos (x))= \ln (2).$$
$$x=\frac{\pi}{4}$$ is a good choice. Then $$s= ?$$.