Question

For \(n=0,1,2, \ldots\), and \(x\) real, prove that $$ |\sin n x| \leq n|\sin x| . $$ Note that this inequality may be false for other values of \(n\). For instance, $$ \mid \sin \langle\pi|>||\sin \pi| . $$

Short answer

Based on the step-by-step solution, answer the following question: Question: Prove the inequality \(|\sin nx| \leq n|\sin x|\) for \(n = 0, 1, 2, \ldots\) and for all real values of \(x\). Answer: The inequality \(|\sin nx| \leq n|\sin x|\) was proven using induction. 1. When \(n = 0\), the inequality holds true as both sides equal 0. 2. Assuming the inequality holds for \(n = k\), we can show that it also holds for \(n = k + 1\) by applying the sum-to-product formula and the Triangle Inequality. 3. Splitting the inequality into two inequalities and simplifying, we see that both of these inequalities hold. 4. The inequality holds for \(n = k + 1\), and thus it is proven by induction for all non-negative integers \(n\). However, it's important to note that the inequality may not hold true for some values of \(n\), as it is only valid for non-negative integers (i.e., \(n = 0, 1, 2, \ldots\)).

Step-by-step solution

Case n = 0

When \(n = 0\), the inequality becomes: $$ |\sin (0 \cdot x)| \leq 0|\sin x| $$ Since both sides of the inequality are equal to 0 (as \(\sin 0 = 0\)), the inequality holds true for \(n = 0\).

Induction

Now let's assume that the inequality holds for some \(n = k\). That is, $$ |\sin kx| \leq k|\sin x| $$ We need to show that this inequality holds for \(n = k + 1\). So, let's multiply both sides of the inequality with \((k + 1)\): $$ |k\sin x| |\sin(kx + x)| \leq (k + 1)|\sin x| $$ We can use the sum-to-product formula \(\sin(A + B) = \sin A \cos B + \cos A \sin B\). In our case, \(A = kx\) and \(B = x\). Thus, we get: $$ | k\sin x | |\sin kx \cos x + \cos kx \sin x| \leq (k + 1)|\sin x| $$

Using the Triangle Inequality

Now, we can use the Triangle Inequality which is \(|a + b| \leq |a| + |b|\) for any real numbers \(a\) and \(b\). So, we have: $$ | k\sin x | |\sin kx \cos x + \cos kx \sin x| \leq | k\sin x | (|\sin kx \cos x| + |\cos kx \sin x|) $$

Splitting the Inequality

Now, we can split the inequality as follows, for the two terms: $$ | k\sin x | |\sin kx \cos x| \leq k|\sin x| |\sin kx \cos x| $$ and $$ | k\sin x | |\cos kx \sin x| \leq k|\sin x| |\cos kx \sin x| $$

Simplification

Notice that, in both of these inequalities, we can divide both sides by the common factor \(|k\sin x|\). This would yield the following inequalities: $$ |\sin kx \cos x| \leq |\sin kx \cos x| $$ $$ |\cos kx \sin x| \leq |\cos kx \sin x| $$ Since both of the initial inequalities hold true, their sum is also true, which implies that: $$ (k+1) |\sin x| \geq |k\sin x| (|\sin kx \cos x| + |\cos kx \sin x|) $$ Therefore, the inequality holds for \(n = k + 1\).

Induction Proof

Since the inequality holds for \(n = 0\) and if the inequality holds for \(n = k\), it also holds for \(n = k + 1\), we have proven the inequality by induction: $$ |\sin nx| \leq n|\sin x| . $$

Inequality may not hold for some values of n

As mentioned in the exercise note, this inequality may be false for some values of \(n\), such as when: $$ |\sin\pi| > |\sin \pi| $$ This is because the inequality we have proven is only valid for \(n = 0, 1, 2, \ldots\) and may not hold true for other values of \(n\).

Key concepts

Inequality Proof

Understanding how to prove inequalities is a fundamental skill in mathematical analysis. An inequality, like an equation, asserts a relationship between two expressions; however, rather than equality, it indicates that one side is greater or less than another. To prove a proposed inequality like \( |\sin nx| \leq n|\sin x| \), one might use direct methods, or employ the principles of mathematical induction as was done in the provided solution.

In the initial assessment (Step 1), the case when \( n = 0 \) is considered a base case. Checking that both sides of the inequality equal 0 immediately confirms the base case. After establishing the base, induction steps (Step 2 to Step 6) strategically build on an assumption that the inequality holds for a certain \( n = k \), and then deduce that it must also hold for \( n = k + 1 \). This process necessitates meticulous work with the algebraic form of the inequality, keeping a close eye on the absolute value's properties and the potential equality within it.

The Triangle Inequality, a powerful tool mentioned in steps 3 and 4, further simplifies the work. It ensures that the absolute value of the sum of any two numbers is less than or equal to the sum of their absolute values, a principle applied in conjunction with trigonometric identities to uphold the structural integrity of the proof. In Step 5, simplification by eliminating common factors brings the proof towards its conclusion, demonstrating that the inequality holds as assumed in the induction hypothesis.

Mathematical Induction

The principle of mathematical induction is a proof technique used to establish that a given property holds for all natural numbers. It involves two critical steps: proving the base case, and then showing that if the statement is true for an arbitrary case \( n = k \), it must also be true for \( n = k + 1 \). The strength of induction lies in its simplicity and elegance, establishing an infinite chain of truths from a single seed of truth.

In the exercise at hand, this powerful tool was utilized in steps 2 through 6. Once the truth of the inequality for \( n = 0 \) was established, it acted as a domino causing a chain reaction ensuring its validity for each subsequent natural number. Simplification in subsequent steps solidifies the inductive step - the assumption does indeed carry to the next instance. As with most inductive proofs, careful attention to implications and logical consistency is crucial for a successful argument.

An improvement advice for those struggling with induction could include a more in-depth examination of the significance of the base case, and why it is essential to not only show the property for \( n = k + 1 \) but also to ensure that every step in the chain is linked to its predecessor, forming an unbreakable inductive bridge.

Trigonometric Inequalities

Trigonometric inequalities involve expressions with trigonometric functions such as sine, cosine, and tangent, where one side of the inequality is not equal to the other. Solving these inequalities requires a firm grasp of both trigonometric identities and the properties of inequalities. A key identity used in this proof is the sum-to-product formula (as mentioned in Step 3), which expresses the sine of a sum as a product of sine and cosine of two angles.

When dealing with trigonometric inequalities, one must navigate through the intricacies of trigonometric functions' behavior, including their boundedness (that is, all of their values lie between -1 and 1) and periodicity. Such careful attention is demonstrated in the exercise's solution through the application of absolute values and the aforementioned Triangle Inequality. Students may find aid in visualizing the trigonometric functions on the unit circle to comprehend the rationale behind these inequalities.

While the given inequality \( |\sin nx| \leq n|\sin x| \) holds for natural numbers, the note in Step 7 reminds us of its limitation by highlighting an instance when the conditions are not met. This caveat serves as an important reminder that trigonometric inequalities can be particularly sensitive to the domain considered, an area often requiring improvement in understanding for many students.