Thank you for your solutions to this problem to Annie from
Newstead Wood, Andre from Tudor Vianu National College, Romania and
Ben who did not give the name of his school. This is Ben's
solution.

(1) \begin{eqnarray} f(x)&=& x - \sin x \\ f(0)
&=& 0 \\ f'(x) &=& 1-\cos x \geq 0\end{eqnarray} As
$ \cos x \leq 1$ then $f'(x)$ 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 $x \geq 0$,
$\sin x \leq x.$

(2)\begin{eqnarray}f(x) &=& \cos x - \left(1 -
{x^2\over 2}\right)\\ f(0) &=& 0 \\ f'(x) &=& -\sin
x + x \geq 0 \quad \rm {by \ (1)}\end{eqnarray} We have already
shown that $\sin x \leq x$, therefore $x - \sin x \geq 0$. So the
function $f(x) = \cos x - \left(1 - {x^2\over 2}\right)$ is always
increasing when $x\geq 0$ and $f(0)=0$. For this to be true $\cos x
\geq 1 - {x^2\over 2}$ when $x \geq 0$.

(3) \begin{eqnarray} f(x) &=& \left(x - {x^3 \over
3!}\right) - \sin x \\ f(0) &=& 0 \\ f'(x) &=& 1 -
{x^2\over 2} - \cos x \leq 0 \quad \rm{by\ (2)} \end{eqnarray} As
$\cos x \geq 1 - {x^2\over 2}$ then $f'(x)$ here must always be
negative. Therefore $f(x)=\left(x - {x^3 \over 3!}\right) - \sin x
$ is always decreasing for $x\geq 0$. As $f(0)= 0$, and the
function decreases as $x$ increases, then $f(x)\leq 0$ when $x \geq
0$. Hence $\sin x \geq \left(x - {x^3 \over 3!}\right)$ for $x \geq
0$.

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

(4) \begin{eqnarray}f(x) &=& \cos x - \left(1 - {x^2
\over 2!} + {x^4\over 4!}\right) \\ f(0)&=&0 \\
f'(x)&=& - \sin x + x - {x^3\over 3!}\leq 0 \quad \rm{by\
(3)} \end{eqnarray} As $\sin x \geq \left(x - {x^3 \over
3!}\right)$ then the function $f(x) = \cos x - \left(1 - {x^2 \over
2!} + {x^4\over 4!}\right)$ is always decreasing, and again $f(0) =
0$ so $f(x)\leq 0$ for $x\geq 0$. If this is true then $\cos x \leq
\left(1 - {x^2\over 2!} + {x^4 \over 4!}\right)$ for $x \geq
0$.

(5) If we continue this process, we see that $$1 - {x^2\over
2!} + {x^4\over 4!} - {x^6\over 6!} \leq \cos x \leq 1 - {x^2\over
2!} + {x^4\over 4!}$$ 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) $$\cos x = 1 -
{x^2\over 2!}+ {x^4\over 4!} -{x^6\over 6!} + ...$$ 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) $$\sin x
= x - {x^3\over 3!}+ {x^5\over 5!} -{x^7\over 7!} + ...$$

A Maclaurin
series is a Taylor series
expansion of a function about $0$.

The mathematicians
Brook Taylor and
Colin Maclaurin were contemporaries who both developed Newton's
work on calculus.