Problem 64

Question

Show that for any numbers $a$ and $b,$ the sine inequality $|\sin b-\sin a| \leq|b-a|$ is true.

Step-by-Step Solution

Verified

Answer

Using the Mean Value Theorem for sine shows that $|\sin b - \sin a| \leq |b - a|$.

1Step 1: Understanding the Problem

We need to demonstrate that the absolute difference between the sine of two angles, $a$ and $b$, is less than or equal to the difference between those angles in absolute terms. In formula terms, we want to prove that $|\sin b - \sin a| \leq |b - a|$.

2Step 2: Use the Mean Value Theorem

According to the Mean Value Theorem, for a continuous function like $\sin x$, there exists a $c$ between $a$ and $b$ such that $\sin b - \sin a = (b - a)\cos c$.

3Step 3: Apply the Property of Cosine

Since the maximum absolute value of $\cos c$ is $1$, we know that $|(b - a)\cos c| \leq |b - a| \cdot 1 = |b - a|$.

4Step 4: Conclude the Proof

Key Concepts

Mean Value TheoremTrigonometric FunctionsInequality ProofCosine Property

Mean Value Theorem

The Mean Value Theorem (MVT) is a fundamental concept in calculus that helps us understand how a function behaves between two points. For any continuous and differentiable function over an interval, the theorem states that there exists at least one point where the derivative (or slope) of the function is equal to the average rate of change over that interval.

In mathematical terms, for a function $ f(x) $, if $ a $ and $ b $ lie within the domain, there exists a $ c $ such that:

\[ f'(c) = \frac{f(b)-f(a)}{b-a} \]This implies that the instantaneous rate of change at $ c $ will match the average rate over $ [a, b] $.
For the sine function, a continuous and differentiable trigonometric function, MVT implies that there is some $ c $ in $ (a, b) $ making the equation $ \sin b - \sin a = (b-a)\cos c $ hold true. This provides a powerful tool for proving inequalities like the sine inequality.

Trigonometric Functions

Trigonometric functions like sine and cosine are periodic functions that relate angles to side lengths in right-angled triangles. They are fundamental in mathematics for modeling waveforms, circular motion, and much more.

The sine function $ \sin(x) $ gives the y-coordinate of a point on the unit circle.
The cosine function $ \cos(x) $ provides the x-coordinate of that point.

Both sine and cosine oscillate between -1 and 1, making their absolute values always less than or equal to 1. This property is essential in understanding how these functions behave and aids in deriving various inequalities, such as the one presented in our exercise.

Inequality Proof

Inequality proofs in mathematics are crucial for establishing boundaries and limits of functions. For the sine inequality stated in this exercise, we aim to prove that the difference in values of sine at two points is bounded by the actual interval between those points.
To do this, we applied the Mean Value Theorem which showed that:\[ \sin b - \sin a = (b-a)\cos c \]Given that $ |\cos c| \leq 1 $, we deduced that:

$|(b-a)\cos c| \leq |b-a| \cdot 1 = |b-a|$.
This led to $|\sin b - \sin a| \leq |b-a|$.

Thus, this clear step-by-step approach closes the loop in proving our sine inequality.

Cosine Property

The cosine function has unique characteristics that are very useful in mathematical proofs, especially for inequalities. One of the key properties is that the cosine of an angle in a unit circle ranges between -1 and 1.
This bound is instrumental when we evaluate expressions involving cosine, as it limits the maximum effect that multiplication by $ \cos(x) $ can have on any given value.

With $ \cos(c) \, \, \epsilon \,[ -1, 1] $, it's guaranteed that $ |\cos c| \leq 1 $.
This limitation allows us to derive inequalities in trigonometric expressions, ensuring they are always valid, like our proven inequality: $ |(b-a)\cos c| \leq |b-a| $.

By leveraging this bound on the cosine function, we can efficiently manage and resolve many mathematical problems involving trigonometric functions.

Problem 64

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