Inequality proof

[Imum Coeli]

Question:
Prove abs(cos(sin(x_1))-cos(sin(x_2))) <= abs(x_1-x_2) for all x_1, x_2 that belong to the reals.

Notes.
I haven't gotten anywhere. I've tried using all the trig identities I know to rewrite it but I can't see anything helpful. I was wondering if I could use the Mean Value Theorem and if so, how? If anyone could point me in the right direction that would be great.

 

[daon2]

Question:
Prove abs(cos(sin(x_1))-cos(sin(x_2))) <= abs(x_1-x_2) for all x_1, x_2 that belong to the reals.

Notes.
I haven't gotten anywhere. I've tried using all the trig identities I know to rewrite it but I can't see anything helpful. I was wondering if I could use the Mean Value Theorem and if so, how? If anyone could point me in the right direction that would be great.

Have you tried:

\(\displaystyle \sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A-B}{2}\right) = -\dfrac{1}{2}[\cos(A)-\cos(B)]\)

And maybe make use of (If you can use it): \(\displaystyle \sin(x) \le x\) for \(\displaystyle x\ge 0\)

 

[Imum Coeli]

Sorry I can't read LaTex very well.

I think that top line is sin((A+B)/2)*sin((A-B)/2) = -1/2*(cos(A)-cos(B)) where A = sin(x_1) and B = sin(x_2).

I have written it like that but I still can't seem to see anything...

Surely as this is part of a calculus course then I should use calculus. That's why I was thinking of the mean value theorem or something. I just can't see what to use where.

 

[daon2]

What strikes me is the following:

Let \(\displaystyle F(x) = \cos(sin(x))\).

Then your proof is equivalent to proving (for \(\displaystyle x\neq y\) which is handled easily by itself):

\(\displaystyle \left|\dfrac{F(x)-F(y)}{x-y}\right| \le 1\)

That should be a big shout out to the mean value theorem.

 

[Imum Coeli]

Yes that looks good! Thanks! I'm a little bit busy with other stuff now but I will work on that as soon as I can. Thanks heaps.

**EDIT
(Sorry if this gets messy. I have no idea if this is right.)

So we have abs(cos(sin(x_1))-cos(sin(x_2))) <= abs(x_1-x_2) for all x_1, x_2 that belong to the reals.

Now
0 <= abs(cos(sin(x_1))-cos(sin(x_2))) <= 2
and
0 <= abs(x_1-x_2) <= infinity
so abs(cos(sin(x_1))-cos(sin(x_2))) <= abs(x_1-x_2) is true for all x >= 2

Now as pointed out above the question becomes
abs((F(x_1) - F(x_2)) / (x_1 - x_2)) <= 1 on [0,2]
so then by the mean value theorem
f'(c) <= 1 on [0,2]

Now fix x_2 = 0 and x_1 = 2
so
f'(c) = 1/2*abs(F(2) -1)
but as
0 <= abs(F(2) -1) <= 2
we have
f'(c) <= 1 on [0,2]

Therefore abs(cos(sin(x_1))-cos(sin(x_2))) <= abs(x_1-x_2) for all x_1, x_2 that belong to the reals. QED

 

[daon2]

Yes that looks good! Thanks! I'm a little bit busy with other stuff now but I will work on that as soon as I can. Thanks heaps.

**EDIT
(Sorry if this gets messy. I have no idea if this is right.)

So we have abs(cos(sin(x_1))-cos(sin(x_2))) <= abs(x_1-x_2) for all x_1, x_2 that belong to the reals.

Now
0 <= abs(cos(sin(x_1))-cos(sin(x_2))) <= 2
and
0 <= abs(x_1-x_2) <= infinity
so abs(cos(sin(x_1))-cos(sin(x_2))) <= abs(x_1-x_2) is true for all x >= 2

Now as pointed out above the question becomes
abs((F(x_1) - F(x_2)) / (x_1 - x_2)) <= 1 on [0,2]
so then by the mean value theorem
f'(c) <= 1 on [0,2]

Now fix x_2 = 0 and x_1 = 2
so
f'(c) = 1/2*abs(F(2) -1)
but as
0 <= abs(F(2) -1) <= 2
we have
f'(c) <= 1 on [0,2]

Therefore abs(cos(sin(x_1))-cos(sin(x_2))) <= abs(x_1-x_2) for all x_1, x_2 that belong to the reals. QED

It looks like you're headed in the right direction. One issue I see is that you are taking the derivative of an absolute value (it looks like you're trying anyway?), which can fail.

Instead: Without loss of generality assume \(\displaystyle x_1>x_2\) and let \(\displaystyle F(x)=\cos(\sin(x))\).

I'll do case 1: \(\displaystyle F(x_1) > F(x_2)\).

By the MVT, since \(\displaystyle F(x)\) is differentiable on \(\displaystyle (x_2,x_1)\) and continuous on \(\displaystyle [x_2,x_1]\), there is a \(\displaystyle c\in (x_2,x_1)\) such that

\(\displaystyle \left|\dfrac{F(x_1)-F(x_2)}{x_1-x_2}\right| = \dfrac{F(x_1)-F(x_2)}{x_1-x_2} = F'(c) \le |F'(c)|\)

But \(\displaystyle |F'(c)| = |-\sin(\sin(c))\cos(c)| = |\sin(\sin(c))|\cdot |\cos(c)| \le 1\cdot 1 = 1\).

Get it?