You are asked in part (5) about generalising this method. What formulae would you expect to derive this way? This method shows a sequence of polynomial approximations to the trig. functions. You are not being asked to give a rigorous proof that the formulae hold in general; that requires a little more work.

Here is another route to the same result. It is not a solution to the problem because the problem specified another method. This should help in seeing the bigger picture more clearly.

We know $\cos x \leq 1$ for all $x$ .
The function $f_{1} (x) = 1 - \cos x$ is positive for all x and so the integral of this function from 0 to $x$ is positive for all $x$ . Hence

$\int_{0}^{x} f_{1} (x) dx = x - \sin x$

is positive so $\sin x \leq x$ .

Integrating again, where $f_{2} (x) = x - \sin x$

$\int_{0}^{x} f_{2} (x) dx = \frac{x^{2}}{2} + \cos x - 1$

is positive so $\cos x \geq 1 - \frac{x^{2}}{2}$ .

Integrating again, where $f_{3} (x) = \cos x - (1 - \frac{x^{2}}{2})$

$\int_{0}^{x} f_{3} (x) dx = \sin x - (x - \frac{x^{3}}{3!})$

is positive so $\sin x \geq x - \frac{x^{3}}{3!}$ .

Integrating again, where $f_{4} (x) = \sin x - (x - \frac{x^{3}}{3!})$

$\int_{0}^{x} f_{4} (x) dx = - \cos x - (\frac{x^{2}}{2!} - \frac{x^{4}}{4!}) + 1$

is positive so $\cos x \leq 1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!}$

and so on .....