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How to prove that the function $\sin x$ is a lipschitz continuous function on the real line using the definition of lipschitz continuity. I know $|\frac{df}{dx}|\leq 1.$ But I wanted to prove without using this condition.

Robert Z
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thomus
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1 Answers1

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Use this identity (see here): $$\sin(x)-\sin(y)=2\cos((x+y)/2)\sin((x-y)/2).$$ Hence $$|\sin(x)-\sin(y)|\leq 2|\sin((x-y)/2)|\leq |x-y|.$$ where we used the inequalities $|\cos t|\leq 1$ and $|\sin t|\leq |t|$.

Robert Z
  • 145,942