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 cosx £ 1 for all x.
The function f1(x) = 1 - cosx is positive for all x and so the integral of this function from 0 to x is positive for all x. Hence
ó
õ 		x

0 
 f1 (x) dx = x -sinx
is positive so sinx £ x.

Integrating again, where f2(x) = x - sinx
ó
õ 		x

0 
 f2(x) dx = x2
2
 + cosx - 1
is positive so
cosx ³ 1 - x2
2

.

Integrating again, where
f3(x) = cosx - (1 - x2
2
 )


ó
õ 		x

0 
 f3(x) dx = sinx - (x - x3
3!
 )
is positive so
sinx ³ x- x3
3!

.

Integrating again, where
f4(x) = sinx - (x - x3
3!
 )


ó
õ 		x

0 
 f4(x) dx = -cosx - ( x2
2!
 - x4
4!
 )+ 1
is positive so
cosx £ 1 - x2
2!
 + x4
4!

and so on .....