29.2 Prove that |cos x – cos y| ≤ |x – y| for all x, y in R.
Solution:
In order to prove the above statement, we need to prove first some trigonometric
basic formulas.
1)
Let’s prove the identity:
xy
xy
sin
2
2
( 1)
cos( x  y )  cos x cos y  sin x sin y
( 2)
cos( x  y )  cos x cos y  sin x sin y
( 3)
cos x  cos y  2 sin
Proof:
We start from the basic formulas:
By subtracting first equality from the second one, we will have
cos( x  y )  cos( x  y )  2 sin x sin y
( 4)
We denote now
xy 
xy 
( 5)
By adding the 2 equations, one yields
2x    

x

2
( 6)
Now, we subtract the second equation from the first one:
2y    

y 
 
2
( 7)
Introducing in (4), we will obtain:
cos   cos   2 sin

 
sin
2
2
( 8)
or, by changing the sign:
cos   cos   2 sin

 
sin
2
2
( 9)
and the proof of (1) is complete (the only thing is that we have to change the
notations from () to (x, y) ).
2)
Let’s prove now that
sin(x)  x , () x > 0
( 10)
Proof:
We will use a geometric judgement.
Let’s consider a circle of radius R = 1.
R=1
A
L =  (rad)

B C
L = sin
It is easy to see that the length of the segment AB is L = R.sin = sin (since R =
1), while the length of the arc AC is L = R. =  (rad).
As the figure shows, we will have always
AB < AC 
sin < .
and the proof is done. (the arc AC can be compared like an hypotenuse in the
triangle ABC, which is always bigger than the cathetes).
Remark: There is no other proof using calculus or other sophisticated rules.
Instead, the above inequality is used to prove some theorems in Calculus.
The graph of below shows the relative positions of “sine” curve and “line” y = x.
y
Y=x
Y = sinx
1
O

2
x
-1
It is not hard to see that y = sin(x) is always located under the line y = x.
(But this is not a proof!)
In order to extend (1) for all the real axis (that is including negative numbers, we
can write (1) as
sin x  x
Now, we can use this last inequality and (9) to prove our statement:
( 11)
cos x  cos y  2 sin
xy
xy
xy xy
sin
 2 sin

2
2
2
2
xy
 x  y sin
2
( 12)
But, as the graph shows us, we have always
sin x  1

sin
xy
1
2
By replacing this in (12), we get finally our equality:
cos x  cos y  x  y
and this is valid for all x, y  R (see (13)).
The proof is now complete.
I hope this is the answer that you are looking for.
If additional explanations are needed, let me know. Good luck!
( 13)