<s>[INST] How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$? 
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.

Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots$$
So, $$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
However I'm stucked from here on. Thank you in advance!
 [/INST] Since you only need the terms up to and including $x^4$, it would be easier just to use the Binomial theorem and get$$\cos^6x=\left[1+\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)\right]^6$$
$$=1+6\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)+15\left(-\frac{x^2}{2}+\frac{x^4}{24}+...\right)^2+...$$
$$=1-3x^2+4x^4...$$
</s>