Alphabeta Math
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Further Trigonometric Identities and Inverse Functions: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-03Open item page →

Machin's formula π/4=4arctan(1/5)arctan(1/239)\pi/4=4\arctan(1/5)-\arctan(1/239)

Example

Machin's formula is

π4=4arctan15arctan1239.\frac\pi4=4\arctan\frac15-\arctan\frac1{239}.

Facts & Assumptions

Given: No hypotheses beyond those quantified in the statement.

[L2]

The tangent addition and subtraction formulas hold when their displayed denominators and domains are nonzero (Addition and subtraction formulas for tangent, cotangent, secant, and cosecant on their exact domains).

Proof

technique · direct
1.1

Put a:=arctan(1/5)a:=\arctan(1/5) and b:=arctan(1/239)b:=\arctan(1/239). Then tana=1/5\tan a=1/5 and tanb=1/239\tan b=1/239. Comparing with tan0=0\tan0=0 and [L3] on the principal branch gives 0<a,b<π/40<a,b<\pi/4.

L1L3
2.1

Since 0<a<π/40<a<\pi/4, we have 0<2a<π/20<2a<\pi/2, so the tangent addition formula applies. It gives tan(2a)=512<1=tan(π/4).\tan(2a)=\frac5{12}<1=\tan(\pi/4). Strict increase of tangent on the principal branch now gives 2a<π/42a<\pi/4, hence 0<4a<π/20<4a<\pi/2.

step 1.1L1L2L3algebra
3.1

The addition formula, now applied to 2a+2a2a+2a, gives tan(4a)=120119>1.\tan(4a)=\frac{120}{119}>1. All displayed denominators are positive.

step 2.1L2algebra
4.1

A final use of [L2] gives tan(4ab)=120/1191/2391+(120/119)(1/239)=1.\tan(4a-b)= \frac{120/119-1/239}{1+(120/119)(1/239)}=1. The denominator is positive.

step 1.1step 3.1L2algebra
4.2

By step 2.1, 4a4a lies in the principal tangent interval. Since tan(4a)>1=tan(π/4)\tan(4a)>1=\tan(\pi/4), strict increase gives 4a>π/4>b4a>\pi/4>b. Thus 0<4ab<4a<π/20<4a-b<4a<\pi/2. Consequently 4ab4a-b and π/4\pi/4 lie in the same injective branch of tangent.

step 1.1step 2.1step 3.1L1L3
5.1

Steps 4.1 and 4.2 imply 4ab=π/44a-b=\pi/4, which is the claimed formula.

step 4.1step 4.2
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-03Open item page →

arcsin(sinx)\arcsin(\sin x) is not the identity outside the principal interval

Example

The identity arcsin(sinx)=x\arcsin(\sin x)=x is not valid for every xRx\in\mathbb R. For example,

arcsin(sin(3π/4))=π/43π/4.\arcsin(\sin(3\pi/4))=\pi/4\ne3\pi/4.

Facts & Assumptions

Given: No hypotheses beyond those quantified in the statement.

[L1]

Principal arcsine is the inverse of sine with values restricted to [π/2,π/2][-\pi/2,\pi/2] (Principal inverse sine and inverse cosine).

[L2]

The supplementary identity is sin(πx)=sinx\sin(\pi-x)=\sin x (Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions).

Proof

technique · direct
1.1

By [L2], sin(3π/4)=sin(π/4)\sin(3\pi/4)=\sin(\pi/4). Since π/4\pi/4 lies in the principal range of arcsine, [L1] gives arcsin(sin(3π/4))=π/4\arcsin(\sin(3\pi/4))=\pi/4.

L1L2algebra
2.1

Since π/43π/4\pi/4\ne3\pi/4, step 1.1 is the claimed counterexample.

step 1.1algebra
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-03Open item page →

Principal arcsine has no finite derivative at 1-1 or 11

Example

The principal arcsine arcsin:[1,1][π/2,π/2]\arcsin:[-1,1]\to[-\pi/2,\pi/2] has no finite derivative at either endpoint 11 or 1-1 (with the library's relative, one-sided endpoint convention).

Facts & Assumptions

Given: No hypotheses beyond those quantified in the statement.

[L1]

On [1,1][-1,1], sin(arcsiny)=y\sin(\arcsin y)=y, and arcsin(1)=π/2\arcsin(1)=\pi/2, arcsin(1)=π/2\arcsin(-1)=-\pi/2 (Principal inverse sine and inverse cosine).

[L3]

(sinx)=cosx(\sin x)'=\cos x, while cos(π/2)=cos(π/2)=0\cos(\pi/2)=\cos(-\pi/2)=0 (The derivatives of sine and cosine are cosine and minus sine, Quarter-turn values and shifts by pi/2 and pi, Pythagorean and parity identities for all six trigonometric functions on their natural domains).

Proof

technique · contradiction
1.1

Suppose arcsin\arcsin had a finite derivative at 11. Differentiate the identity sin(arcsiny)=y\sin(\arcsin y)=y at 11 relative to [1,1][-1,1]. The derivative of the right side is 11, whereas [L2] and [L3] make the derivative of the left side cos(π/2)(arcsin)(1)=0\cos(\pi/2)(\arcsin)'(1)=0, a contradiction.

assume-contraL1L2L3
1.2

The identical argument at 1-1 gives 1=cos(π/2)(arcsin)(1)=01=\cos(-\pi/2)(\arcsin)'(-1)=0.

assume-contraL1L2L3
2.1

Therefore neither finite endpoint derivative exists.

step 1.1step 1.2discharge-contradiction

Sources