TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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 $x$ , $\sin x=0\Longleftrightarrow x=m\pi\text{ for some }m\in\mathbb Z,\qquad \cos x=0\Longleftrightarrow x=(m+1/2)\pi\text{ for some }m\in\mathbb Z.$ Both sine and cosine have period $2\pi$ , and no smaller positive number is a common period.

Facts & Assumptions

Given: A real $x$ .

[L1]

$\pi$ is the first positive sine zero, and sine is positive on $(0,\pi)$ (Pi is the first positive zero of sine).

[L2]

Shifts by $\pi$ negate sine and cosine, while shifts by $\pi/2$ 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 $m \le x < m + 1$ , Integer powers $a^m$ , The principle of mathematical induction, Laws of integer exponents).

Proof

technique · direct

1.1

Natural induction applied to the $\pi$ shift gives $\sin(x+n\pi)=(-1)^n\sin x$ and $\cos(x+n\pi)=(-1)^n\cos x$ for every natural $n$ . Applying these identities at $x-n\pi$ 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 $m\pi\le x<(m+1)\pi$ and put $r=x-m\pi\in[0,\pi)$ . By [L1] and step 1.1, $\sin x=0$ exactly when $r=0$ , hence exactly when $x=m\pi$ .

L1step 1.1L3

3.1

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

step 2.1L2

4.1

Step 1.1 with $m=2$ gives period $2\pi$ . A positive common period $T$ has $\sin T=0$ , hence $T=m\pi$ by step 2.1; $m=1$ fails for cosine because $\cos\pi=-1$ , so $m\ge2$ .

step 2.1step 3.1L1L2

5.1

Therefore $2\pi$ is the least positive common period.

step 4.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 87 results over 23 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

NIST Digital Library of Mathematical Functions, Chapter 4 (standard reference, not scraped)
C. Schmeiser, Introduction to Analysis (standard reference, not scraped)