<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] $$\cos x=\frac {e^{ix}+e^{-ix}}2$$
so
$$32\cos^6x=\cos 6x+6\cos 4x+15\cos 2x+10=$$$$=(1-18x^2+54x^4+\dots)+(6-48x^2+64x^4-\dots)+(15-30x^2+10x^4-\dots)+10$$
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