Answer
${\cos ^2}x = \dfrac{1}{2} + \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{2^{2n - 1}}{x^{2n}}}}{{\left( {2n} \right)!}}$,
converges for all $x$.
Work Step by Step
From Table 2, we get $\cos x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}$, converges for all $x$.
Substituting $2x$ for $x$ in the series above gives
$\cos 2x = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{{\left( {2x} \right)}^{2n}}}}{{\left( {2n} \right)!}} = \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{2^{2n}}{x^{2n}}}}{{\left( {2n} \right)!}}$
Since ${\cos ^2}x = \dfrac{1}{2}\left( {1 + \cos 2x} \right)$, so
${\cos ^2}x = \dfrac{1}{2}\left( {1 + \cos 2x} \right) = \dfrac{1}{2}\left( {1 + \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{2^{2n}}{x^{2n}}}}{{\left( {2n} \right)!}}} \right)$
${\cos ^2}x = \dfrac{1}{2} + \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{2^{2n - 1}}{x^{2n}}}}{{\left( {2n} \right)!}}$
