Answer
$\cos^2 x=1-x^2+\dfrac{x^4}{3}-\dfrac{2x^6}{45}-....$
Work Step by Step
The Taylor series can be written as follows: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-...........$ and $\cos x=1-\dfrac{x^2}{2}+\dfrac{ x^3}{3}-....$
We have: $\cos^2 x=\cos 2x +\sin^2 x$
or, $=(1-\dfrac{(2x)^2}{2}+\dfrac{ (2x)^3}{3}-....) + (\dfrac{2x^2}{2!}-\dfrac{2^3 x^4}{4!}+....)$
$\implies =1-\dfrac{2x^2}{2!}+\dfrac{(2)^3 x^4}{4!}-\dfrac{(2)^5 x^6}{6!}-$
$ \implies =1-x^2+\dfrac{x^4}{3}-\dfrac{2x^6}{45}-........$
$\implies \cos^2 x=1-x^2+\dfrac{x^4}{3}-\dfrac{2 \space x^6}{45}-....$
