LemmaStatement: AI-adaptedProof: AI-generatedprecheck 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 $0 < x \leq 2$ , $\sin x \geq x - x^{3} / 6 \geq x / 3 > 0$ . Also $\cos 2 \leq - 1 / 3$ . Consequently $\cos$ is strictly decreasing on $[0, 2]$ .

Facts & Assumptions

Given: A real $x$ with $0 < x \leq 2$ .

[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

1.1L1L2algebra

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

1.2L1L2algebra

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 $\cos 2 \leq - 1 / 3$ .

2.1step 1.1L3∎

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

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources