Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Parity and the Pythagorean identity for sine and cosine

Statement

For every real x, sin⁡(−x)=−sin⁡x,cos⁡(−x)=cos⁡x,sin⁡2x+cos⁡2x=1. Consequently ∣sin⁡x∣≤1 and ∣cos⁡x∣≤1.

Facts & Assumptions

Proof

technique · direct
1.1

The derivative of F(x)=sin⁡2x+cos⁡2x is 2sin⁡xcos⁡x−2cos⁡xsin⁡x=0.

L2L3algebra
2.1

Step 1.1 makes F differentiable everywhere, hence continuous by [L4]. The zero-derivative theorem therefore makes F constant, and F(0)=1, proving sin⁡2x+cos⁡2x=1; each square is then at most 1.

step 1.1L2L3L4algebra
3.1

Applying the addition formulas at x+(−x)=0 gives 0=sin⁡xcos⁡(−x)+cos⁡xsin⁡(−x),1=cos⁡xcos⁡(−x)−sin⁡xsin⁡(−x). The coefficient matrix squares to the identity by step 2.1, so these equations give cos⁡(−x)=cos⁡x and sin⁡(−x)=−sin⁡x.

L1L2step 2.1algebra∎

Depends on

Used by

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Dependency tree · two levels

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Sources