Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02
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The zero sets of sine and cosine and the least positive common period 2 pi

Statement

For every real x, sin⁡x=0⟺x=mπ for some m∈Z,cos⁡x=0⟺x=(m+1/2)π for some m∈Z. Both sine and cosine have period 2π, and no smaller positive number is a common period.

Facts & Assumptions

Given: A real x.

[L1]

π is the first positive sine zero, and sine is positive on (0,π) (Pi is the first positive zero of sine).

[L2]

Shifts by π negate sine and cosine, while shifts by π/2 exchange them up to sign (Quarter-turn values and shifts by pi/2 and pi).

[L3]

Every real has an integer part; every integer is a natural number or the negative of a natural number; natural induction and the integer-power laws are valid (Integer part: for every real x there is exactly one integer m with m≤x<m+1, Integer powers am, The principle of mathematical induction, Laws of integer exponents).

Proof

technique · direct
1.1

Natural induction applied to the π shift gives sin⁡(x+nπ)=(−1)nsin⁡x and cos⁡(x+nπ)=(−1)ncos⁡x for every natural n. Applying these identities at x−nπ gives the matching backward shifts; since every integer is n or −n and (−1)−n=(−1)n, the displayed identities hold for every integer m.

L2L3algebra
2.1

Choose an integer m with mπ≤x<(m+1)π and put r=x−mπ∈[0,π). By [L1] and step 1.1, sin⁡x=0 exactly when r=0, hence exactly when x=mπ.

L1step 1.1L3
3.1

The quarter-turn shift converts the sine zero set into cos⁡x=0 exactly when x=(m+1/2)π.

step 2.1L2
4.1

Step 1.1 with m=2 gives period 2π. A positive common period T has sin⁡T=0, hence T=mπ by step 2.1; m=1 fails for cosine because cos⁡π=−1, so m≥2.

step 2.1step 3.1L1L2
5.1

Therefore 2π is the least positive common period.

step 4.1∎

Depends on

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Sources