(1)

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =x-\sin x\\ \multicolumn{1}{c}{f(0)}& =0\\ \multicolumn{1}{c}{f\text{'}(x)}& =1-\cos x\ge 0.\end{array}}$

As $\cos x\le 1$ then $f\text{'}(x)$ is always positive so the function $f(x)=x-\sin x$ is always increasing, so $x$ is becoming increasingly larger than sin x and therefore, when $x\ge 0$, $\sin x\le x.$

(2)

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =\cos x-(1-\frac{x^2}{2})\\ \multicolumn{1}{c}{f(0)}& =0\\ \multicolumn{1}{c}{f\text{'}(x)}& =-\sin x+x\ge 0\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{by}\hspace{0.5em}(1)\end{array}}$

We have already shown that $\sin x\le x$, therefore $x-\sin x\ge 0$. So the function $f(x)=\cos x-(1-\frac{x^2}{2})$ is always increasing when $x\ge 0$ and $f(0)=0$. For this to be true $\cos x\ge 1-\frac{x^2}{2}$ when $x\ge 0$.

(3)

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =(x-\frac{x^3}{3!})-\sin x\\ \multicolumn{1}{c}{f(0)}& =0\\ \multicolumn{1}{c}{f\text{'}(x)}& =1-\frac{x^2}{2}-\cos x\le 0\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{by}\hspace{0.5em}(2)\end{array}}$

As $\cos x\ge 1-\frac{x^2}{2}$ then $f\text{'}(x)$ here must always be negative. Therefore $f(x)=(x-\frac{x^3}{3!})-\sin x$ is always decreasing for $x\ge 0$. As $f(0)=0$, and the function decreases as $x$ increases, then $f(x)\le 0$ when $x\ge 0$. Hence $\sin x\ge (x-\frac{x^3}{3!})$ for $x\ge 0$.

Note that the derivative of $\frac{x^n}{n!}$ is $\frac{{\mathrm{nx}}^{n-1}}{n!}=\frac{x^{n-1}}{(n-1)!}$ and we can write $2=2!$.

(4)

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{f(x)}& =\cos x-(1-\frac{x^2}{2!}+\frac{x^4}{4!})\\ \multicolumn{1}{c}{f(0)}& =0\\ \multicolumn{1}{c}{f\text{'}(x)}& =-\sin x+x-\frac{x^3}{3!}\le 0\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{by}\hspace{0.5em}(3)\end{array}}$

As $\sin x\ge (x-\frac{x^3}{3!})$ then the function $f(x)=\cos x-(1-\frac{x^2}{2!}+\frac{x^4}{4!})$ is always decreasing, and again $f(0)=0$ so $f(x)\le 0$ for $x\ge 0$. If this is true then $\cos x\le (1-\frac{x^2}{2!}+\frac{x^4}{4!})$ for $x\ge 0$.

(5) If we continue this process, we see that

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

As we introduce more terms this series gets closer to $\cos x$.

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

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

By a similar look at the terms for $\sin x$, we get the infinite Maclaurin series for $\sin x$, again valid for all $x$ (in radians)

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