<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] $$f\left( x \right) =f\left( 0 \right) +{ f\left( 0 \right)  }^{ \prime  }x+\frac { { f\left( 0 \right)  }^{ \prime \prime  } }{ 2! } { x }^{ 2 }+\frac { { f\left( 0 \right)  }^{ \prime \prime \prime  } }{ 3! } { x }^{ 3 }+\frac { { f\left( 0 \right)  }^{ \prime \prime \prime \prime  } }{ 4! } { x }^{ 4 }+...\\ f\left( x \right) =\cos ^{ 6 }{ x } \\ f\left( 0 \right) =1\\ f^{ \prime  }\left( x \right) =-6\cos ^{ 5 }{ x } \sin { x } \Rightarrow f^{ \prime  }\left( 0 \right) =-6\cos ^{ 5 }{ 0 } \sin { 0 } =0\\ f^{ \prime \prime  }\left( x \right) ={ \left( -6\cos ^{ 5 }{ x } \sin { x }  \right)  }^{ \prime  }=30\cos ^{ 4 }{ x } \sin ^{ 2 }{ x-6\cos ^{ 6 }{ x }  } \Rightarrow f^{ \prime \prime  }\left( 0 \right) =-6\\ f^{ \prime \prime \prime  }\left( x \right) ={ \left( 30\cos ^{ 4 }{ x } \sin ^{ 2 }{ x-6\cos ^{ 6 }{ x }  }  \right)  }^{ \prime  }=-120\cos ^{ 3 }{ x } \sin ^{ 3 }{ x } +60\cos ^{ 5 }{ x } \sin { x } +36\cos ^{ 5 }{ x } \sin { x } \Rightarrow f^{ \prime \prime \prime  }\left( 0 \right) =0\\ f^{ \prime \prime \prime \prime  }\left( x \right) ={ \left( -120\cos ^{ 3 }{ x } \sin ^{ 3 }{ x } +60\cos ^{ 5 }{ x } \sin { x } +36\cos ^{ 5 }{ x } \sin { x }  \right)  }^{ \prime  }=360\cos ^{ 2 }{ x } \sin ^{ 4 }{ x } -360\sin ^{ 2 }{ x } \cos ^{ 4 }{ x } -300\cos ^{ 4 }{ x } \sin ^{ 2 }{ x } +60\cos ^{ 6 }{ x } -180\cos ^{ 4 }{ x } \sin ^{ 2 }{ x } +36\cos ^{ 5 }{ x } \Rightarrow \\ f^{ \prime \prime \prime \prime  }\left( 0 \right) =96\\ \cos ^{ 6 }{ x } =1-\frac { 6 }{ 2! } { x }^{ 2 }+\frac { 96 }{ 4! } { x }^{ 4 }+...=1-3{ x }^{ 2 }+4{ x }^{ 4 }+..\\ $$
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