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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.

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

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