Question

Upper Bounds Show that for any numbers \(a\) and \(b\) \(|\sin b-\sin a| \leq|b-a|\)

Short answer

The inequality \(|\sin b - \sin a| \leq |b-a|\) can be proved by expressing the left hand side as the difference of two sine functions and applying Mean Value Theorem(MVT). According to MVT, there exists a number c in the interval (a, b) such that the derivative of the function at c is equal to the average rate of change of function over the interval. Using this theorem and the fact that the absolute value of the cosine function is always less than or equal to 1, the inequality is proved.

Step-by-step solution

Express the problem as a difference of two functions

We can express the left hand side of the inequality as the difference of two sine functions: \(f(b) - f(a)\) where \(f(x) = \sin x\). Now, we should think about how we can use the Mean Value Theorem (MVT) to prove the inequality.

Apply the Mean Value Theorem (MVT)

According to the MVT, there exists a number \(c\) in the interval \((a, b)\) such that \(f'(c) = (f(b) - f(a)) / (b - a)\) where \(f'(x)\) is the derivative of \(f(x)\). So, \(f'(c) = \cos c\) and \(f'(c) = (\sin b - \sin a) / (b - a)\).

Find Absolute values and Apply Inequality

We know that the absolute value of the derivative of f(x), which is \(\cos x\) is always less than or equal to 1. So, \(|\cos c| \leq 1\). From the previous step, we know that \(|\cos c| = |(\sin b - \sin a) / (b - a)|\). So, we get, \(|(\sin b - \sin a) / (b - a)| \leq 1\).

Multiply both sides by |b-a|

Multiply both sides by \(|b-a|\) to get \(|\sin b - \sin a| \leq |b - a|\). This completes the proof