Finding the Maclaurin Expansion for \(f(x) =\cos x\)

Find the Maclaurin expansion for \(f(x) =\cos x.\)

Solution We apply the differentiation property of power series to the Maclaurin expansion for \(\sin x.\) \[ \begin{eqnarray*} \frac{d}{dx}\sin x &=& \frac{d}{dx}\!\left( x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!} -\cdots +(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}+\cdots \right) \\[5pt] &=&\frac{d}{dx}\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k+1}}{ (2k+1) !} \end{eqnarray*} \]

Then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \cos x=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\cdots +(-1)^{n}\frac{x^{2n}}{(2n)!}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\,(-1) ^{k}\frac{x^{2k}}{(2k) !}}} \]

for all numbers \(x\).