The Generalised Sine Inequality

This problem is a masterclass in proof architecture. Rather than attacking $|\sin(nx)| \le n|\sin x|$ directly with induction — which forces you to wrestle with trigonometric identities and induction logic simultaneously — the scaffold decouples the two: Part (i) isolates all the trigonometry into a single lemma, and Part (ii) turns induction into pure algebra. Parts (iii) and (iv) then show how one general result unlocks a family of corollaries.

Problem Statement

Let $x_1, x_2, \dots, x_n$ be real numbers, and let $n \in \mathbb{N}$ with $n \ge 1$.

i. Prove that for any two real numbers $x_1$ and $x_2$:

$$|\sin(x_1 + x_2)| \le |\sin x_1| + |\sin x_2|$$

ii. Use mathematical induction, or otherwise, to prove that for any $n \ge 1$:

$$\left|\sin\!\left(\sum_{i=1}^n x_i\right)\right| \le \sum_{i=1}^n |\sin x_i|$$

iii. Deduce that for any $x \in \mathbb{R}$ and $n \in \mathbb{N}$:

$$|\sin(nx)| \le n|\sin x|$$

iv. Using the results above, prove that for any $x \in \mathbb{R}$ and $n \in \mathbb{N}$:

$$1 - \cos(nx) \le n^2(1 - \cos x)$$

---

Hints

• Part (i): Use the angle addition formula for sine, apply the triangle inequality ($|a + b| \le |a| + |b|$), then bound the cosine factors using $|\cos\theta| \le 1$.
• Part (ii): For the induction step ($k \Rightarrow k+1$), group the first $k$ terms as a single angle $X = x_1 + \dots + x_k$, and apply the lemma from Part (i) to $(X + x_{k+1})$.
• Part (iii): Set $x_1 = x_2 = \dots = x_n = x$ in the result from Part (ii).
• Part (iv): Use the half-angle identity $1 - \cos(2\theta) = 2\sin^2\theta$. Write $1 - \cos(nx)$ in this form, then apply Part (iii) and square the resulting inequality.

---

Solutions

Part (i)

Using the angle addition formula and the triangle inequality:

$$|\sin(x_1 + x_2)| = |\sin x_1 \cos x_2 + \cos x_1 \sin x_2| \le |\sin x_1||\cos x_2| + |\cos x_1||\sin x_2|$$

Since $|\cos\theta| \le 1$ for all real $\theta$:

$$|\sin(x_1 + x_2)| \le |\sin x_1|(1) + (1)|\sin x_2| = |\sin x_1| + |\sin x_2| \qquad \square$$

Part (ii)

Base case ($n = 1$): $|\sin(x_1)| \le |\sin x_1|$ holds trivially. (Note: the case $n = 2$ is exactly Part (i).)

Inductive step: Suppose the inequality holds for some $k \ge 1$, i.e.

$$\left|\sin\!\left(\sum_{i=1}^k x_i\right)\right| \le \sum_{i=1}^k |\sin x_i|$$

We must show it holds for $k + 1$. Write:

$$\left|\sin\!\left(\sum_{i=1}^{k+1} x_i\right)\right| = \left|\sin\!\left(\left(\sum_{i=1}^k x_i\right) + x_{k+1}\right)\right|$$

Treat $\sum_{i=1}^k x_i$ as a single angle and apply the lemma from Part (i):

$$\le \left|\sin\!\left(\sum_{i=1}^k x_i\right)\right| + |\sin x_{k+1}|$$

Applying the inductive hypothesis to the first term:

$$\le \left(\sum_{i=1}^k |\sin x_i|\right) + |\sin x_{k+1}| = \sum_{i=1}^{k+1} |\sin x_i| \qquad \square$$

Note how the induction step contains no trigonometry at all — all the trig work was encapsulated in the lemma from Part (i).

Part (iii)

Set $x_i = x$ for all $i \in \{1, 2, \dots, n\}$ in Part (ii):

$$\left|\sin\!\left(\sum_{i=1}^n x\right)\right| \le \sum_{i=1}^n |\sin x|$$ $$|\sin(nx)| \le n|\sin x| \qquad \square$$

Part (iv)

Recall the half-angle identity $1 - \cos(2\theta) = 2\sin^2\theta$. Setting $2\theta = nx$, i.e. $\theta = \dfrac{nx}{2}$:

$$1 - \cos(nx) = 2\sin^2\!\left(\frac{nx}{2}\right)$$

From Part (iii) applied with $x \mapsto \frac{x}{2}$:

$$\left|\sin\!\left(n \cdot \frac{x}{2}\right)\right| \le n\left|\sin\!\left(\frac{x}{2}\right)\right|$$

Since both sides are non-negative, squaring preserves the inequality:

$$\sin^2\!\left(\frac{nx}{2}\right) \le n^2 \sin^2\!\left(\frac{x}{2}\right)$$

Multiply both sides by $2$:

$$2\sin^2\!\left(\frac{nx}{2}\right) \le 2n^2 \sin^2\!\left(\frac{x}{2}\right)$$

Substituting the half-angle identity back on both sides:

$$1 - \cos(nx) \le n^2(1 - \cos x) \qquad \square$$

---

Takeaways

• The Paradox of Generalisation: Proving the harder, more general statement (a sum of $n$ different angles) is logically cleaner than proving the specific case directly. Generalising decouples the trigonometry from the induction logic, making both steps simpler.
• Modularity in Proofs: Breaking a proof into lemmas creates reusable "black boxes." In Part (ii), the induction step is pure algebra — there are no sines or cosines in sight. The trigonometric heavy-lifting was done once in Part (i) and then simply invoked.
• Translating Between Sine and Cosine: Inequalities bounding $|\sin|$ can often be translated into inequalities bounding cosine via the Pythagorean identity or half-angle formulas. Recognising that $1 - \cos\theta = 2\sin^2(\theta/2)$ is the key bridge in Part (iv).

---

Further Readings

HSC Induction, HSC Sequences