Thomas' Calculus 13th Edition

$\dfrac{1}{2}$
Consider the Maclaurin Series for $\cos x$ as follows: $\cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}=1-\dfrac{1}{2!}x^2+\dfrac{1}{4!}x^4-...$ As we are given that $x=\dfrac{\pi}{3}$ Then, we have $\cos (x)=\cos (\dfrac{\pi}{3})=\dfrac{1}{2}$