Solutions - 0250-02

Write the alternating series for $\cos (0.02)$:

$$\begin{array}{c}\cos (x)=\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n}}{(2n)!}\\ \\ \gg \gg \cos (0.02)=\sum _{n=0}^{\infty }(-1)\frac{{(0.02)}^{2n}}{(2n)!}\\ \\ \gg \gg 1-\frac{{(0.02)}^2}{2!}+\frac{{(0.02)}^4}{4!}-\frac{{(0.02)}^6}{6!}\end{array}$$

This is an alternating series, so we can apply the “Next Term Bound” rule. Calculate some terms:

$$\frac{{(0.02)}^2}{2!}=2.0\times {10}^{-4},\frac{{(0.02)}^4}{4!}=6.7\times {10}^{-9}$$

(Without a calculator, we can see that ${(0.02)}^4=2^4\times {10}^{-8}=1.6\times {10}^{-7}$. Dividing by $4!$ will only decrease this value.)

So we add up the prior terms:

$$\gg \gg 1-\frac{{(0.02)}^2}{2!}=0.9998$$