Mean value theorem(mvt) to prove the inequality

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The discussion revolves around using the Mean Value Theorem (MVT) to prove the inequality |sin a - sin b| ≤ |a - b| for all a and b. Participants explore the MVT's conditions of continuity and differentiability, noting challenges with absolute values in the context of inequalities. The MVT is applied by stating that f'(c) = (sin b - sin a) / (b - a) = cos(c), leading to the conclusion that |cos(c)| ≤ 1. This implies that |sin a - sin b| ≤ |a - b| holds true. The conversation emphasizes the importance of understanding the application of MVT in proving inequalities.
Khayyam89
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Homework Statement


Essentially, the question asks to use the mean value theorem(mvt) to prove the inequality: \left|sin a -sin b \right| \leq \left| a - b\right| for all a and b


The Attempt at a Solution



I do not have a graphing calculator nor can I use one for this problem, so I need to prove that the inequality basically by proof. What I did was to look at the mvt hypotheses: if the function is continuous and differetiable on closed and open on interval a,b, respectively. However, the problem I am having is that I am getting thrown off by the absolute values and the fact that I've never used mvt on inequalities. I know the absolute value of the sin will look like a sequence of upside-down cups with vertical tangents between them. Hints most appreciated.
 
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The mean value theorem states that in an interval [a,b]:

f'(c) = \frac{ \sin b - \sin a}{b-a} = \cos(c)

Now put absolute value signs there and make use of |\cos(x)| \leq 1
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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