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CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Parity and the Pythagorean identity for sine and cosine

Statement

For every real xx, sin(x)=sinx,cos(x)=cosx,sin2x+cos2x=1.\sin(-x)=-\sin x,\qquad\cos(-x)=\cos x,\qquad\sin^2x+\cos^2x=1. Consequently sinx1|\sin x|\le1 and cosx1|\cos x|\le1.

Facts & Assumptions

Proof

technique · direct
1.1

The derivative of F(x)=sin2x+cos2xF(x)=\sin^2x+\cos^2x is 2sinxcosx2cosxsinx=02\sin x\cos x-2\cos x\sin x=0.

L2L3algebra
2.1

Step 1.1 makes FF differentiable everywhere, hence continuous by [L4]. The zero-derivative theorem therefore makes FF constant, and F(0)=1F(0)=1, proving sin2x+cos2x=1\sin^2x+\cos^2x=1; each square is then at most 11.

step 1.1L2L3L4algebra
3.1

Applying the addition formulas at x+(x)=0x+(-x)=0 gives 0=sinxcos(x)+cosxsin(x),1=cosxcos(x)sinxsin(x).0=\sin x\cos(-x)+\cos x\sin(-x),\qquad 1=\cos x\cos(-x)-\sin x\sin(-x). The coefficient matrix squares to the identity by step 2.1, so these equations give cos(x)=cosx\cos(-x)=\cos x and sin(x)=sinx\sin(-x)=-\sin x.

L1L2step 2.1algebra

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 59 results over 22 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources