(1)

f(x) = x - sinx f(0) = 0 f¢(x) = 1-cos x ³ 0 .

As cosx £ 1 then f¢(x) is always positive so the function f(x) = x - sinx is always increasing, so x is becoming increasingly larger than sin x and therefore, when x ³ 0, sinx £ x.

(2)

f(x) = cosx - (1 - x2 2 ) f(0) = 0 f¢(x) = -sinx + x ³ 0     by  (1)

We have already shown that sinx £ x, therefore x - sinx ³ 0. So the function

f(x) = cosx - (1 -

x2 2

)

is always increasing when x ³ 0 and f(0)=0. For this to be true

cosx ³ 1 -

x2 2

when x ³ 0.

(3)

f(x) = (x - x3 3! ) - sinx f(0) = 0 f¢(x) = 1 - x2 2 - cosx £ 0     by (2)

As

cosx ³ 1 -

x2 2

then f¢(x) here must always be negative. Therefore

f(x)=(x -

x3 3!

) - sinx

is always decreasing for x ³ 0. As f(0) = 0, and the function decreases as x increases, then f(x) £ 0 when x ³ 0. Hence

sinx ³ (x -

x3 3!

)

for x ³ 0.

Note that the derivative of

xn n!

is

nxn-1 n!

=

xn-1 (n-1)!

and we can write 2=2!.

(4)

f(x) = cosx - (1 - x2 2! + x4 4! ) f(0) =0 f¢(x) = - sinx + x - x3 3! £ 0     by (3)

As

sinx ³ (x -

x3 3!

)

then the function

f(x) = cosx - (1 -

x2 2!

+

x4 4!

)

is always decreasing, and again f(0) = 0 so f(x) £ 0 for x ³ 0. If this is true then

cosx £ (1 -

x2 2!

+

x4 4!

)

for x ³ 0.

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

1 - x2 2! + x4 4! - x6 6! £ cosx £ 1 - x2 2! + x4 4! .

As we introduce more terms this series gets closer to cosx.

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

cosx = 1 - x2 2! + x4 4! - x6 6! + ...

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

sinx = x - x3 3! + x5 5! - x7 7! + ...