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!

I think the simplest way is by expanding
to get terms with degrees below 4 we have to choose which combination of terms of original cos(x) series must be chosen and sum up the posibble combination
here is the answer:
$$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
$$=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)\times\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)\times...\times\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)$$
so there is this combinations:
1-$$choosing \; all \; 1's \; from \; 6 \; parentheses \;that\;can \;be \;done \;in \;one \;uniqe \;way\; so \; term \;1 \;is \;produced :\;1$$
2-$$choosing \; 1's \; from \; 5 \; parentheses \; and- \frac{1}{2}x^2 \; or \; \frac{1}{24}x^4 from \;other \;parentheses \;which \;can \;be \;done \;in\; 6 \;defferent \;ways\; so \; terms \; - \frac{6}{2}x^2 and \frac{6}{24}x^4\;are \;produced :\;- 3\times x^2 + \frac{1}{4}x^4$$
3-$$choosing \; 1's \; from \; 4 \; parentheses \; and- \frac{1}{2}x^2 \;  from\; 4\;other \;parentheses \;which \;can \;be \;done \;in\; \binom{6}{2} \;defferent \;ways\; so \; term \;  \frac{15}{4}x^4 is \;produced :\;+ \frac{15}{4}x^4$$
4-there is no other term possible with degree below 6. summing up above terms results in the truncated series as below:
$$1 + 3\times x^2 + 4\times x^4 + O(x^6)$$
