LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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Sine is positive and cosine is strictly decreasing on (0,2), with cos 2 at most -1/3

Statement

For $0<x\le2$ , $\sin x\ge x-x^3/6\ge x/3>0$ . Also $\cos2\le-1/3$ . Consequently $\cos$ is strictly decreasing on $[0,2]$ .

Facts & Assumptions

Given: A real $x$ with $0<x\le2$ .

[L1]

The defining sine and cosine series are those of Sine and cosine defined by their real power series.

[L2]

An alternating series with decreasing nonnegative terms has its sum between consecutive partial sums (The alternating series test: if $(b_k)$ is nonincreasing with $b_k \to 0$ then $\sum_{k} (-1)^{k} b_k$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_n$ ).

[L3]

$\cos'=-\sin$ , and the mean value theorem detects strict monotonicity from derivative sign (The derivatives of sine and cosine are cosine and minus sine, The mean value theorem, as the case $g(x) = x$ of Cauchy's: for $f$ continuous on $[a,b]$ with $a < b$ and differentiable on $(a,b)$ there is $c \in (a,b)$ with $f(b) - f(a) = f'(c)(b-a)$ ).

Proof

technique · direct

1.1

The absolute sine terms after the first have successive ratio at most $4/6<1$ , so [L2] gives $\sin x\ge x-x^3/6\ge x/3>0$ .

L1L2algebra

1.2

In the cosine series at $2$ , the first three terms sum to $1-2+2/3=-1/3$ , and the remaining alternating tail begins negative with decreasing absolute terms; hence $\cos2\le-1/3$ .

L1L2algebra

2.1

On $(0,2)$ one has $\cos'=-\sin<0$ by step 1.1, so the mean value theorem makes $\cos$ strictly decreasing on $[0,2]$ .

step 1.1L3∎

Depends on

Used by

Pi is the first positive zero of sine Corollary
The Bartle-Sherbert bounds 2.828 < pi < 3.185 Example
Cosine has a smallest positive zero, lying strictly between zero and two Theorem
Quarter-turn values and shifts by pi/2 and pi Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 78 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

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