TITLE: Maclaurin polynomial of $\ln(\cos(x))$
QUESTION [5 upvotes]: I want to write down $\ln(\cos(x))$  Maclaurin polynomial of degree 6. I'm having trouble understanding what I need to do, let alone explain why it's true rigorously.
The known expansions of $\ln(1+x)$ and $\cos(x)$ gives:
$$\forall x \gt -1,\ \ln(1+x)=\sum_{n=1}^{k} (-1)^{n-1}\frac{x^n}{n} + R_{k}(x)=x-\frac{x^2}{2}+\frac{x^3}{3}+R_{3}(x)$$
$$\cos(x)=\sum_{n=0}^{k} (-1)^{n}\frac{x^{2n}}{(2n)!} + T_{2k}(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+T_{4}(x)$$
Writing $\ln(1+x)$ with $t=x-1$ gives:
$$\forall t \gt 0,\ \ln(t)=\sum_{n=1}^{k} (-1)^{n-1}\frac{(t+1)^n}{n} + R_{k}(t)$$
But now I'm clueless. 

Do I just 'plug' $\cos(x)$ expansion in $\ln(t)$? Can I even do that?
Isn't it a problem that $\ln(x)$ isn't defined for $x\leq 0$ but $|\cos(x)| \leq 1$?

REPLY [2 votes]: I think Joe Johnson's is a good idea:
$$
\frac{d}{dx} \ln\cos x = -\tan x
$$
plus the knowledge of the  Taylor series for the tangent,
$$
\tan x = \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{(2n)!}x^{2n-1} \ , \qquad \text{for} \qquad x \in (-\pi/2 , \pi/2)
$$
(here, $B_s$ are the Bernouilli numbers) gives you everything:
$$
\ln\cos x = C - \int \tan x dx = C - \int \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{(2n)!}x^{2n-1} dx = C - \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{2n(2n)!}x^{2n}
$$
Now, taking $x = 0$ on both sides, you get $C = 0$, so
$$
\ln\cos x = - \sum_{n=1}^\infty \frac{B_{2n}(-4)^n(1-4^n)}{2n(2n)!}x^{2n} \ , \qquad \text{for} \qquad x \in (-\pi/2 , \pi/2) \ .
$$