How do you write the Maclaurin Series for \[\cos {{\left( x \right)}^{2}}?\]

Hint: We are asked to find the Maclaurin Series for \[\cos {{\left( x \right)}^{2}}.\] We will learn what is Maclaurin series and we will learn how to write it. Then using the expansion form of the Maclaurin series \[f\left( x \right)=\sum{\dfrac{{{f}^{n}}\left( 0 \right)}{n!}}{{x}^{n}}\] we will find the derivative of f(x) at x = 0. In our function, \[{{\cos }^{2}}\left( x \right)\] we will first simplify it using \[{{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}\] and then we will find the various derivatives. Once we have the derivative then we will compare and form the required series.

Complete step by step answer:
We are given a function \[\cos {{\left( x \right)}^{2}}\] and it is nothing but \[{{\cos }^{2}}x.\] We have to find the Maclaurin Series of \[{{\cos }^{2}}x.\] To start solving our problem we will understand what does this series mean and how it is formed. Now, Maclaurin Series is a special type of Taylor series expansion which is just expanded about x = 0. As we know Taylor Series of any function say f(x) is given as
\[f\left( x \right)=f\left( a \right)+{{f}^{'}}\left( a \right)+\dfrac{{{f}^{''}}\left( a \right)}{2!}.....\]
Or
\[f\left( x \right)=\sum\limits_{n=0}^{n=\infty }{\dfrac{{{f}^{n}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}}\]
where a is the point of expansion.
Since in Maclaurin, the point of expansion is zero only. So Maclaurin series will be given as
\[f\left( x \right)=\sum\limits_{n=0}^{n=\infty }{\dfrac{{{f}^{n}}\left( 0 \right)}{n!}{{x}^{n}}}\]
where \[{{f}^{n}}\left( 0 \right)\] means the derivative of the function at 0.
Now, we have the function defined as \[{{\cos }^{2}}\left( x \right)\] and as it is a 2-degree function, its derivative will be lengthy and we will convert it into linear. We know \[{{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}\] and so by definition of Maclaurin Series, we need to find the derivative of our function \[{{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}\] around x = 0. So, we will find the first 4 derivatives of \[f\left( x \right)=\dfrac{1+\cos 2x}{2}.\]
First derivative
\[{{f}^{'}}\left( x \right)=\dfrac{d\left( \dfrac{1+\cos 2x}{2} \right)}{dx}\]
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{0+2\left[ -\sin 2x \right]}{2}\]
\[\Rightarrow {{f}^{'}}\left( x \right)=-\sin 2x\]
At x = 0, we have
\[\Rightarrow {{f}^{'}}\left( 0 \right)=-\sin \left( 2\times 0 \right)\]
\[\Rightarrow {{f}^{'}}\left( 0 \right)=-\sin 0\]
\[\Rightarrow {{f}^{'}}\left( 0 \right)=0\]
Second derivative
\[{{f}^{'}}\left( x \right)=-\sin \left( 2x \right)\]
\[{{f}^{''}}\left( x \right)=-\left[ \cos 2x \right].2\]
\[\Rightarrow {{f}^{''}}\left( x \right)=-2\cos 2x\]
At x = 0, we have,
\[\Rightarrow {{f}^{''}}\left( x \right)=-2\left( \cos 0 \right)\]
\[\Rightarrow {{f}^{''}}\left( x \right)=-2\]
Now, similarly the third derivative at x = 0, we will get,
\[{{f}^{'''}}\left( x \right)=4\sin 2x\]
\[\Rightarrow {{f}^{'''}}\left( 0 \right)=0\]
Lastly, the fourth derivative of f(x) at x = 0
\[{{f}^{''''}}\left( x \right)=8\cos 2x\]
\[\Rightarrow {{f}^{''''}}\left( 0 \right)=8\]
Now, we will observe that the odd derivatives are all zero and all even derivatives are 2 to the power of one less than the order of the derivative with the alternating positive and negative sign.
So, putting these values in f(x) we get,
\[f\left( x \right)=f\left( 0 \right)+\dfrac{{{f}^{'}}\left( 0 \right)}{1!}x+\dfrac{{{f}^{''}}\left( 0 \right)}{2!}{{x}^{2}}.....\]
\[\Rightarrow f\left( x \right)=\dfrac{1+\cos \left( 0 \right)}{2}+\dfrac{0\left( x \right)}{1!}+\dfrac{-2{{x}^{2}}}{2!}+\dfrac{0{{x}^{3}}}{3!}+\dfrac{8{{x}^{4}}}{4!}....\]
So,
\[\Rightarrow {{\cos }^{2}}x=1-\dfrac{-2{{x}^{2}}}{2!}+\dfrac{8{{x}^{4}}}{4!}-\dfrac{32{{x}^{6}}}{6!}....\]
Hence our Maclaurin Series is given as
\[{{\cos }^{2}}x=1-\dfrac{-2{{x}^{2}}}{2!}+\dfrac{8{{x}^{4}}}{4!}-\dfrac{32{{x}^{6}}}{6!}....\]
In summation form, we get,
\[{{\cos }^{2}}x=1+\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}\dfrac{{{2}^{2k-1}}}{\left( 2k \right)!}.{{x}^{2k}}}\]

Note: While going for finding the derivative, remember that we have to find the multiple derivatives that are first order, second order, third order and more. So, it is crucial to change the given function to the simplest form. Else our calculation becomes lengthy and mistakes will occur, so we need to be cautious.