Session-authored (Fable 5 assisted)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

✓ 6 results · all verified · 0 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 6 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Fundamental Trigonometric Identities: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

Addition and half-angle identities compute the sine, cosine, and tangent of $\pi/12$

Example

Using $\pi/12=\pi/3-\pi/4$ , $\sin(\pi/12)=\frac{\sqrt6-\sqrt2}{4},\quad \cos(\pi/12)=\frac{\sqrt6+\sqrt2}{4},\quad \tan(\pi/12)=2-\sqrt3.$

Facts & Assumptions

Given: The positive number $\pi$ and the angles $\pi/4,\pi/3\in(0,\pi)$ .

[L1]

Quarter-turn values and shifts by pi/2 and pi gives $\sin(\pi/2)=1$ and $\cos(\pi/2)=0$ .

[L2]

Half-angle identities with the sign determined by the quadrant gives $\cos(x/2)=\varepsilon_c\sqrt{(1+\cos x)/2}$ and $\sin(x/2)=\varepsilon_s\sqrt{(1-\cos x)/2}$ , with the signs of the corresponding half-angle functions.

[L3]

Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions gives $\cos(\pi/2-x)=\sin x$ and $\cos(\pi-x)=-\cos x$ .

[L4]

Double-angle and quadratic power-reduction identities gives $\cos(2x)=2\cos^2x-1$ .

[L5]

Pi is the first positive zero of sine gives $\sin x>0$ whenever $0<x<\pi$ .

[L6]

The subtraction formulas for sine and cosine gives $\sin(u-v)=\sin u\cos v-\cos u\sin v$ and $\cos(u-v)=\cos u\cos v+\sin u\sin v$ .

[L7]

Tangent, cotangent, secant, and cosecant on their exact natural domains defines $\tan x=\sin x/\cos x$ when $\cos x\ne0$ .

Verification

1.1

By [L5], $\sin(\pi/4)>0$ , and [L3] gives $\cos(\pi/4)=\sin(\pi/4)>0$ . Applying [L2] to $x=\pi/2$ and using [L1] therefore gives $\sin(\pi/4)=\cos(\pi/4)=\sqrt2/2$ .

L1L2L3L5algebra

1.2

Put $c=\cos(\pi/3)$ . Then $c>0$ because [L3] writes it as $\sin(\pi/6)$ and [L5] makes that value positive. Also [L3] and [L4] give $-c=\cos(2\pi/3)=2c^2-1$ , hence $(2c-1)(c+1)=0$ ; positivity rules out $c=-1$ , so $\cos(\pi/3)=1/2$ .

L3L4L5algebra

2.1

From [L3] and step 1.2, $\cos(2\pi/3)=-1/2$ . Since $\sin(\pi/3)>0$ by [L5], [L2] applied to $x=2\pi/3$ gives $\sin(\pi/3)=\sqrt{(1+1/2)/2}=\sqrt3/2$ .

L2L3L5step 1.2algebra

3.1

Substitute steps 1.1, 1.2, and 2.1 into [L6] with $u=\pi/3$ and $v=\pi/4$ . This gives $\sin(\pi/12)=(\sqrt6-\sqrt2)/4$ and $\cos(\pi/12)=(\sqrt6+\sqrt2)/4>0$ .

L6step 1.1step 1.2step 2.1algebra

4.1

By [L7] and step 3.1, $\tan(\pi/12)=(\sqrt6-\sqrt2)/(\sqrt6+\sqrt2)$ ; rationalizing gives $(8-4\sqrt3)/4=2-\sqrt3$ .

L7step 3.1algebra∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Morrie's law: $\cos(\pi/9)\cos(2\pi/9)\cos(4\pi/9)=1/8$

Example

$\cos(\pi/9)\cos(2\pi/9)\cos(4\pi/9)=1/8.$ The conventions and prerequisite facts used below are recorded in Double-angle and quadratic power-reduction identities, Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions, The zero sets of sine and cosine and the least positive common period 2 pi.

Facts & Assumptions

Given: $x=\pi/9$ .

Verification

1.1

Apply $\sin(2t)=2\sin t\cos t$ successively at $t=x$ , $2x$ , and $4x$ . This gives $8\sin x\cos x\cos2x\cos4x=\sin8x$ .

algebra

2.1

Here $8x=\pi-x$ , so the supplementary identity gives $\sin8x=\sin x$ . Since $0<x<\pi$ , the sine-zero characterization gives $\sin x\ne0$ . Cancel it in step 1.1 to obtain $8\cos x\cos2x\cos4x=1$ .

step 1.1given∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

Exact sine and cosine values at $\pi/10$ , $\pi/5$ , and $2\pi/5$

Example

Writing $\varphi=(1+\sqrt5)/2$ , one has $\cos(\pi/5)=\varphi/2,\quad \sin(\pi/10)=(\sqrt5-1)/4,\quad \cos(2\pi/5)=(\sqrt5-1)/4,$ with the complementary sine values determined by positive square roots. The conventions and prerequisite facts used below are recorded in $T_n(\cos\theta)=\cos(n\theta)$ and $U_n(\cos\theta)\sin\theta=\sin((n+1)\theta)$ for every $n\in\mathbb N$ , Double-angle and quadratic power-reduction identities, Square roots exist: a unique $\sqrt{a} \ge 0$ with $(\sqrt{a})^2 = a$ ; the positives are $\{x^2 : x \neq 0\}$ , Signs, monotonicity intervals, and ranges of sine and cosine.

Facts & Assumptions

Given: $5(\pi/5)=\pi$ and the sign ranges on the first quadrant.

Verification

1.1

By the defining recurrence for $T_n$ , $T_5(t)=16t^5-20t^3+5t$ . Put $c=\cos(\pi/5)$ . The multiple-angle identity gives $T_5(c)=\cos\pi=-1$ ; factoring $16c^5-20c^3+5c+1=(c+1)(4c^2-2c-1)^2$ and using $0<c<1$ yields $4c^2-2c-1=0$ .

givenalgebra

2.1

The positive solution is $c=(1+\sqrt5)/4=\varphi/2$ . The double-angle identity gives $\cos(2\pi/5)=2c^2-1=(\sqrt5-1)/4$ , and power reduction at $\pi/10$ gives $\sin(\pi/10)=\sqrt{(1-\cos(\pi/5))/2}=(\sqrt5-1)/4$ . The signs are positive because the angles lie in the first quadrant.

step 1.1algebra∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-02Open item page →

$T_0$ through $T_4$ and $U_0$ through $U_3$ from the Chebyshev recurrences

Example

The recurrences give $T_0=1$ , $T_1=x$ , $T_2=2x^2-1$ , $T_3=4x^3-3x$ , $T_4=8x^4-8x^2+1$ , and $U_0=1$ , $U_1=2x$ , $U_2=4x^2-1$ , $U_3=8x^3-4x$ . The conventions and prerequisite facts used below are recorded in Chebyshev polynomials of the first and second kinds by their three-term recurrences, Degrees and leading coefficients of the Chebyshev polynomials.

Facts & Assumptions

Given: The defining recurrences.

Verification

1.1

Apply the $T_n$ recurrence successively and collect like powers through $T_4$ .

algebra

2.1

Apply the $U_n$ recurrence successively and collect like powers through $U_3$ .

algebra∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

$\sin(3\pi/2)=-1$ shows that the positive square root is not an unconditional half-angle formula

Statement refuted

The unconditional assertion $\sin(x/2)=\sqrt{(1-\cos x)/2}$ for every real $x$ is false. The conventions and prerequisite facts used below are recorded in Half-angle identities with the sign determined by the quadrant, Quarter-turn values and shifts by pi/2 and pi.

Facts & Assumptions

Given: $x=3\pi$ .

Counterexample

1.1

Here $\sin(x/2)=\sin(3\pi/2)=-1$ , while $\sqrt{(1-\cos3\pi)/2}=1$ .

given

2.1

Thus the positive square root loses the required quadrant sign.

algebra∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-02Open item page →

$u=v=\pi/4$ shows that the tangent addition formula cannot omit its domain restrictions

Statement refuted

The tangent addition formula cannot be asserted for all real $u,v$ without domain restrictions. The conventions and prerequisite facts used below are recorded in Addition and subtraction formulas for tangent, cotangent, secant, and cosecant on their exact domains, Tangent, cotangent, secant, and cosecant on their exact natural domains, Quarter-turn values and shifts by pi/2 and pi.

Facts & Assumptions

Given: $u=v=\pi/4$ .

Counterexample

1.1

Both $\tan u$ and $\tan v$ equal $1$ , so the formal right side has denominator $1-1=0$ .

given

2.1

Meanwhile $u+v=\pi/2$ , where tangent itself is undefined.

given∎

Sources

Standard references

Recommended treatments; not extraction sources.

Fundamental Trigonometric Identities: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Addition and half-angle identities compute the sine, cosine, and tangent of π/12\pi/12π/12

Example

Facts & Assumptions

Verification

Morrie's law: cos⁡(π/9)cos⁡(2π/9)cos⁡(4π/9)=1/8\cos(\pi/9)\cos(2\pi/9)\cos(4\pi/9)=1/8cos(π/9)cos(2π/9)cos(4π/9)=1/8

Example

Facts & Assumptions

Verification

Exact sine and cosine values at π/10\pi/10π/10, π/5\pi/5π/5, and 2π/52\pi/52π/5

Example

Facts & Assumptions

Verification

T0T_0T0​ through T4T_4T4​ and U0U_0U0​ through U3U_3U3​ from the Chebyshev recurrences

Example

Facts & Assumptions

Verification

sin⁡(3π/2)=−1\sin(3\pi/2)=-1sin(3π/2)=−1 shows that the positive square root is not an unconditional half-angle formula

Statement refuted

Facts & Assumptions

Counterexample

u=v=π/4u=v=\pi/4u=v=π/4 shows that the tangent addition formula cannot omit its domain restrictions

Statement refuted

Facts & Assumptions

Counterexample

Sources

Standard references

Addition and half-angle identities compute the sine, cosine, and tangent of $\pi/12$

Morrie's law: $\cos(\pi/9)\cos(2\pi/9)\cos(4\pi/9)=1/8$

Exact sine and cosine values at $\pi/10$ , $\pi/5$ , and $2\pi/5$

$T_0$ through $T_4$ and $U_0$ through $U_3$ from the Chebyshev recurrences

$\sin(3\pi/2)=-1$ shows that the positive square root is not an unconditional half-angle formula

$u=v=\pi/4$ shows that the tangent addition formula cannot omit its domain restrictions