Use the Taylor Series given in Key Idea 8.8.1 to verify the given identity: cos(-x)=cos(x)
Key Idea 8.8.1 \[ cosx=\Sigma_{n=0} (-1)^{n} \frac{x^{2n}}{(2n)!} \] First Few terms: \[cos(x)= 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+... \] \[cos(-x)=1-\frac{-x^{2}}{2!}+\frac{-x^{4}}{4!}-\frac{-x^{6}}{6!}= 1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+... \] Which is the same as the series for cos(x), therefore cos(-x)=cos(x)
cos(x) is an even function, thus why it has only even powers (sin(x) is an odd function)
Sources: https://people.math.sc.edu/girardi/m142/handouts/10sTaylorPolySeries.pdf