Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Quarter-turn values and shifts by pi/2 and pi

Statement

For every real xx, sin⁡(x+π/2)=cos⁡x,cos⁡(x+π/2)=−sin⁡x,sin⁡(x+π)=−sin⁡x,cos⁡(x+π)=−cos⁡x.\sin(x+\pi/2)=\cos x,\quad\cos(x+\pi/2)=-\sin x,\quad\sin(x+\pi)=-\sin x,\quad\cos(x+\pi)=-\cos x. In particular, sin⁡(π/2)=1,cos⁡(π/2)=0,sin⁡π=0,cos⁡π=−1.\sin(\pi/2)=1,\quad\cos(\pi/2)=0,\quad\sin\pi=0,\quad\cos\pi=-1.

Facts & Assumptions

Given: A real xx and γ=π/2\gamma=\pi/2.

[L1]

cos⁡γ=0\cos\gamma=0 and sin⁡γ>0\sin\gamma>0 (Pi as twice the smallest positive zero of cosine, Cosine has a smallest positive zero, lying strictly between zero and two, Sine is positive and cosine is strictly decreasing on (0,2), with cos 2 at most -1/3).

[L2]

The addition formulas and Pythagorean identity hold (The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine).

Proof

technique · direct
1.1

From cos⁡2γ+sin⁡2γ=1\cos^2\gamma+\sin^2\gamma=1, cos⁡γ=0\cos\gamma=0, and sin⁡γ>0\sin\gamma>0, one gets sin⁡γ=1\sin\gamma=1.

L1L2
2.1

Substituting γ\gamma into the addition formulas gives sin⁡(x+γ)=cos⁡x\sin(x+\gamma)=\cos x and cos⁡(x+γ)=−sin⁡x\cos(x+\gamma)=-\sin x.

step 1.1L2
3.1

Applying step 2.1 twice gives the shifts by 2γ=π2\gamma=\pi and the listed special values.

step 2.1L1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 53 results over 18 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