Show that $\ln\ 2 \approx \frac{\pi^2}{16}\Bigl(1+\frac{\pi^2}{96}\Bigr)$

Question

Show the Maclaurin Series for $\ln(\cos\ x)=-\frac{1}{2}x^2-\frac{1}{12}x^4 + \ ...$ (1st part of the question already worked out )

Show that $\ln 2 \approx \frac{\pi^2}{16}\Bigl(1+\frac{\pi^2}{96}\Bigr)$

Given $x=\frac{\pi}{4}$

Answer 1

Hint:

Since $\cos(\frac{\pi}{4})=\frac{1}{\sqrt 2}$, you can write $$\ln\left(\frac{1}{\sqrt 2}\right)=\ln(2^{-1/2})=-\frac{1}{2}\ln(2)$$

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Edit for completion:

For small values of $\cos(x)$ we have $$\ln(\cos x)\approx-\frac{1}{2}x^2-\frac{1}{12}x^4\tag{1}$$ When $x=\frac{\pi}{4}$ we get $$-\frac{1}{2}\ln(2)=\ln(\cos \frac{\pi}{4})\approx-\frac{1}{2}\left(\frac{\pi}{4}\right)^2-\frac{1}{12}\left(\frac{\pi}{4}\right)^4\\\ln(2)\approx\frac{\pi^2}{16}\left(1-\frac{\pi^2}{96}\right)$$