<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] Let $A=\cos x.$ We have $\cos 2x=2A^2-1.\quad$$ \cos 3 x= 4A^3-3A.\quad$$ \cos 4x=2\cos^22x-1=8A^4-8A^2+1.\quad$$\cos 6x=2(\cos 3x)^2-1=32A^6-48A^4+18A^2-1.$ From the expressions for $\cos 2x,\; \cos 4x,\;\cos 6x$ we have $32A^6=\cos 6x+6\cos 4x+15\cos 2x+10$. So $\cos^6x=\sum_{n=0}^{\infty}(-1)^nx ^{2n}2^{-5}(6^{2n}+6.4^{2n}+15.2^{2n}+10)/(2n)!.$
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