structural:summarization, structural:expansion, information:extraction
A trigonometric inequality

Induction is often used to prove inequalities. As an example, we prove that | sin n x | ≤ n | sin x | {\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|} for any real number x {\displaystyle x} and natural number n {\displaystyle n}.

At first glance, it may appear that a more general version, | sin n x | ≤ n | sin x | {\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|} for any real numbers n, x {\displaystyle n,x}, could be proven without induction; but the case n = 1 2, x = π {\textstyle n={\frac {1}{2}},\,x=\pi } shows it may be false for non-integer values of n {\displaystyle n}. This suggests we examine the statement specifically for natural values of n {\displaystyle n}, and induction is the readiest tool.

Proposition. For any x ∈ R {\displaystyle x\in \mathbb {R} } and n ∈ N {\displaystyle n\in \mathbb {N} }, | sin n x | ≤ n | sin x | {\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|}.

Proof. Fix an arbitrary real number x {\displaystyle x}, and let P (n) {\displaystyle P(n)} be the statement | sin n x | ≤ n | sin x | {\displaystyle \left|\sin nx\right|\leq n\left|\sin x\right|}. We induce on n {\displaystyle n}.

Base case: The calculation | sin 0 x | = 0 ≤ 0 = 0 | sin x | {\displaystyle \left|\sin 0x\right|=0\leq 0=0\left|\sin x\right|} verifies P (0) {\displaystyle P(0)}.

Induction step: We show the implication P (k) ⟹ P (k + 1) {\displaystyle P(k)\implies P(k+1)} for any natural number k {\displaystyle k}. Assume the induction hypothesis: for a given value n = k ≥ 0 {\displaystyle n=k\geq 0}, the single case P (k) {\displaystyle P(k)} is true. Using the angle addition formula and the triangle inequality, we deduce: | sin (k + 1) x | = | sin k x cos x + sin x cos k x | (angle addition) ≤ | sin k x cos x | + | sin x cos k x | (triangle inequality) = | sin k x | | cos x | + | sin x | | cos k x | ≤ | sin k x | + | sin x | (| cos t | ≤ 1) ≤ k | sin x | + | sin x | (induction hypothesis) = (k + 1) | sin x |. {\displaystyle {\begin{aligned}\left|\sin(k+1)x\right|&=\left|\sin kx\cos x+\sin x\cos kx\right|&&{\text{(angle addition)}}\\&\leq \left|\sin kx\cos x\right|+\left|\sin x\,\cos kx\right|&&{\text{(triangle inequality)}}\\&=\left|\sin kx\right|\left|\cos x\right|+\left|\sin x\right|\left|\cos kx\right|\\&\leq \left|\sin kx\right|+\left|\sin x\right|&&(\left|\cos t\right|\leq 1)\\&\leq k\left|\sin x\right|+\left|\sin x\right|&&{\text{(induction hypothesis}})\\&=(k+1)\left|\sin x\right|.\end{aligned}}}

The inequality between the extreme left-hand and right-hand quantities shows that P (k + 1) {\displaystyle P(k+1)} is true, which completes the induction step.

Conclusion: The proposition P (n) {\displaystyle P(n)} holds for all natural numbers n. {\displaystyle n.} Q.E.D.