Question

calculus proof:
pic attached

Answer 1

Since they're both positive, you could try squaring both sides and expanding. I'm not sure just yet what approach is best.

Answer 2

alright, i'll try that

Answer 3

If you divide both sides by \(|x_1-x_2|\) you get a derivative.

Answer 4

Sort of

Answer 5

You get a difference quotient*

Answer 6

Another thing to consider is the other side must range between 0 and 2.

Answer 7

\[\frac{ |\cos(\sin(x))-\cos(\sin(y))| }{ |x-y| }\le 1\]

Answer 8

like that?

Answer 9

Yes, but I'm just speculating. Differentiation probably won't help.

Answer 10

oh ok. why would the range be 0-2?

Answer 11

Because abolute values are greater than 0 and the highest it could go is to be \(1 - (-1)=2\)

Answer 12

Basically you only need to worry about the case where \[
0\leq|x_1-x_2|\leq 2
\]

Answer 13

x=pi/2 y=pi/2

Answer 14

Are there any theorems that have been introduced lately that could help?

Answer 15

well for this tute, we have learnt mean value theorem, rolle theorem

Answer 16

I figured something like that...

Answer 17

Okay so tell me the mean value theorem.

Answer 18

Something like \(f\) is continuous on \((a,b)\) and there is some \(c\in (a,b)\) where \[
f(b)-f(a)=f'(c)(b-a)
\]

Answer 19

just what i was going to write...haha

Answer 20

wait continous on a closed interval and differentiable on an open interval

Answer 21

Or in other words: \[
\frac{f(b)-f(a)}{b-a} = f'(c)
\]

Answer 22

yep

Answer 23

Maybe let \(f(x)=\cos(\sin(x))\) and differentiate.

Answer 24

We'll chose an interval after the fact.