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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21
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Sine and cosine are 1-Lipschitz on R

Statement

For all real u,v,

∣sin⁡u−sin⁡v∣≤∣u−v∣,

and

∣cos⁡u−cos⁡v∣≤∣u−v∣.

Thus sine and cosine are Lipschitz functions on R, each with Lipschitz constant 1.

Facts & Assumptions

Given: Real numbers u,v and the functions sine and cosine on R.

[L1]

The functions sin⁡ and cos⁡ are differentiable on R, with (sin⁡x)′=cos⁡x and (cos⁡x)′=−sin⁡x (The derivatives of sine and cosine are cosine and minus sine).

[L2]

For every real x, ∣sin⁡x∣≤1 and ∣cos⁡x∣≤1 (Parity and the Pythagorean identity for sine and cosine).

[L3]

A function differentiable on a set is continuous at every point of that set (A function differentiable at c is continuous at c).

[L4]

If I⊆R is order-convex, f:I→R is continuous on I and differentiable at every interior point, and ∣f′(x)∣≤M there for some M≥0, then ∣f(x)−f(y)∣≤M∣x−y∣ for all x,y∈I (If f is continuous on an interval I and ∣f′∣≤M at every interior point, then ∣f(x)−f(y)∣≤M∣x−y∣ for all x,y∈I, so f is Lipschitz with constant M and uniformly continuous on I).

Proof

technique · direct
1.1L1L2L3

By [L1] and [L2], ∣sin⁡′x∣=∣cos⁡x∣≤1 and ∣cos⁡′x∣=∣sin⁡x∣≤1 for every real x; both functions are continuous on R by [L3].

2.1step 1.1L4

Apply [L4] to sine on the order-convex set R with M=1: ∣sin⁡u−sin⁡v∣≤∣u−v∣.

3.1step 1.1L4∎

Apply [L4] to cosine on the same set with M=1: ∣cos⁡u−cos⁡v∣≤∣u−v∣.

Depends on

Dependency tree · 2 levels

Sources