(2)

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =\cos x-(1-\frac{x^2}{2})\\ \multicolumn{1}{c}{f(0)}& =0\\ \multicolumn{1}{c}{f\text{'}(x)}& =-\sin x+x\ge 0\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{by}(1)\end{array}}$

Hence $\cos x\ge 1-\frac{x^2}{2}$ for $x\ge 0$.

(3)

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =(x-\frac{x^3}{3!})-\sin x\\ \multicolumn{1}{c}{f(0)}& =0\\ \multicolumn{1}{c}{f\text{'}(x)}& =1-\frac{x^2}{2}-\cos x\le 0\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{by}(2)\end{array}}$

Hence $\sin x\ge (x-\frac{x^3}{3!})$ for $x\ge 0$.

(4)

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =\cos x-(1-\frac{x^2}{2!}+\frac{x^4}{4!})\\ \multicolumn{1}{c}{f(0)}& =0\\ \multicolumn{1}{c}{f\text{'}(x)}& =-\sin x+x-\frac{x^3}{3!}\le 0\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{by}(3)\end{array}}$

Hence $\cos x\le (1-\frac{x^2}{2!}+\frac{x^4}{4!})$ for $x\ge 0$.

(5) This process can be repeated indefinitely to give the Maclaurin series, valid for all $x$ (in radians)

${\displaystyle \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...}$

${\displaystyle \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...}$