Chapter 5.4, Problem 66E

To determine

(a)

To prove:

By comparison theorem to the inequality $\sin x\le x$ for $x>0$ to show:

$1-\frac{x^2}{2}\le \cos x\le 1$

To determine

(b)

To prove:

By applying comparison theorem again to the inequality $1-\frac{x^2}{2}\le \cos x\le 1$ to show:

$x-\frac{x^3}{6}\le \sin x\le x$, for $x>0$

To determine

(c)

To calculate:

Verify inequalities $1-\frac{x^2}{2}\le \cos x\le 1$ and $x-\frac{x^3}{6}\le \sin x\le x$ for $x=0.3$.