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- Thread starter NWeid1
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- #1

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- #2

gb7nash

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Start with what you know. What does the mean value theorem say?

- #3

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- #4

gb7nash

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(sin(b)-sin(a)) = f'(c)(b-a)

It's a well-known fact that x = y → |x| = |y|.

How can we apply that?

- #5

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I honestly don't know...lol :\

- #6

gb7nash

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x = y → |x| = |y|

This says that:

left-hand side = right-hand side

can be rewritten as:

|left-hand side| = |right-hand side|

For example:

a+b = c+d → |a+b| = |c+d|

Now, let's look at the problem at hand. We have:

(sin(b)-sin(a)) = f'(c)(b-a)

Applying the absolute value on both sides, how can we rewrite this?

- #7

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|sin(b)-sin(a)| = |f'(c)(b-a)|. I understand this, I just read it wrong, lol sorry

- #8

gb7nash

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|sin(b)-sin(a)| = |f'(c)(b-a)|

Ok, good. Now, let's calculate what f'(c) is. What's f'(x)? After you have that, plug c in.

- #9

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f'(x) = cosx so f'(c) = cos(c).

- #10

gb7nash

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f'(x) = cosx so f'(c) = cos(c).

Correct. So now, we have:

|sin(b)-sin(a)| = |cos(c)(b-a)|

It's also a well-known fact that |xy| = |x||y|. How could we rewrite the right-hand side?

- #11

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|sin(b)-sin(a)| = |cos(c)||b-a|. And -1<=cosx<=1 so would it be |sin(b)-sin(a)| <= 1|b-a| ?

- #12

gb7nash

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|sin(b)-sin(a)| = |cos(c)||b-a|. And -1<=cosx<=1 so would it be |sin(b)-sin(a)| <= 1|b-a| ?

Correct.

- #13

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Thank you so much for working with me and not just posting an unexplained proof. Props!

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