arc-equals-cosine

Arc Equals Its Cosine

Summary: Problem I of chapter 22 (§531). On the unit circle, find the arc $s$ such that $s=\cos s$. By false-position-method, $s=42^{\circ }20^{\prime }47^{\prime \prime }14^{\prime \prime \prime }$, arc length $0.7390847$. The same numerical constant resurfaces, via linear translation of the unknown, in Problems III and IV.

Sources: chapter22, §§531, 533, 534.

Last updated: 2026-05-12.

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Statement and existence

On a circle of radius 1, every arc $s\in (0,\pi /2)$ has a cosine. At $s=0$ the cosine is $1>0=s$; at $s=\pi /2$ the cosine is $0<\pi /2=s$. So somewhere strictly between, the two equal each other. Existence is the only structural observation Euler makes (§531) — uniqueness he takes for granted, justified by the strict monotonicity of $s-\cos s$ on that interval.

False-position computation

The bracket-and-interpolate iteration is the canonical illustration of false-position-method:

Bracket coarsely. $s=30^{\circ }$ gives $s=0.5236$, $\cos s=0.8660$ — far apart, $\cos s$ much larger. $s=40^{\circ }$: $s=0.6981$, $\cos s=0.7660$ — still $\cos s>s$. $s=45^{\circ }$: $s=0.7854$, $\cos s=0.7071$ — now flipped. So $40^{\circ }<s<45^{\circ }$.

Interpolate. Errors at $40^{\circ }$ and $45^{\circ }$ are $+403166$ and $-458049$ (in units of the last digit of the log table). The proportion $859215:403166=5^{\circ }:2^{\circ }20^{\prime }$ pushes the estimate past $42^{\circ }$.

Refine. Try $s=42^{\circ }$ and $s=43^{\circ }$:

$$\begin{array}{rcc}s=& 42^{\circ }& 43^{\circ }\\ \log s=& 9.8651267& 9.8753459\\ \log \cos s=& 9.8710735& 9.8641275\\ \text{error}& +0.0059468& -0.0112184\end{array}$$

Total $0.0171652$, giving $\Delta =59468/171652\approx 0^{\circ }20^{\prime }47^{\prime \prime }$ past $42^{\circ }$.

One more pass at $s=2140^{\prime }$ and $s=2141^{\prime }$ refines the answer to seconds: $s=42^{\circ }20^{\prime },47^{\prime \prime },14^{\prime \prime \prime }$.

Final answer.

$$\boxed{s_0=42^{\circ }20^{\prime }47^{\prime \prime }14^{\prime \prime \prime },\text{arc length }=0.7390847.}$$

Why it reappears: III and IV

Two later problems of the chapter reduce to this same constant by a linear substitution.

Problem III (§533): ordinate bisecting a quadrant — see equal-area-bisection-problems

Find $DE$ such that the region $ADE$ (under arc and ordinate) has half the area of the quadrant $ACB$.

Setup: with arc $AE=s$ and the quadrant area $\pi /4$,

$$\frac{1}{2}s-\frac{1}{2}\sin s\cos s=\frac{\pi }{8},$$

i.e., $s-\pi /4=\frac{1}{2}\sin 2s$. Substitute $u=s-\pi /4$, so $2s=\pi /2+2u$ and $\sin 2s=\cos 2u$:

$$u=\frac{1}{2}\cos 2u.$$

This is exactly Problem I in the doubled variable $2u$:

$$2u=\cos (2u),2u=s_0,$$

so $u=s_0/2=21^{\circ }10^{\prime }23^{\prime \prime }37^{\prime \prime \prime \prime }$, and $AE=u+45^{\circ }=66^{\circ }10^{\prime }23^{\prime \prime }37^{\prime \prime \prime \prime }$.

Problem IV (§534): chord bisecting a semicircle — see equal-area-bisection-problems

Find $AD$ in semicircle $AEDB$ such that the two halves of the semicircle have equal area.

Setup: with arc $AD=s$, the sector $ACD=s/2$ and the cut-off segment is $\frac{1}{2}s-\frac{1}{2}\sin s$, equated to half the semicircle $\pi /4$:

$$s-\sin s=\frac{\pi }{2},$$

i.e., $s-\pi /2=\sin s$. Substitute $u=s-\pi /2$, so $\sin s=\cos u$:

$$u=\cos u.$$

This is literally Problem I: $u=s_0=42^{\circ }20^{\prime }47^{\prime \prime }14^{\prime \prime \prime }$, hence $s=u+90^{\circ }=132^{\circ }20^{\prime }47^{\prime \prime }14^{\prime \prime \prime }$.

The reduction map

Problem	Substitution	Reduced equation
I	—	$s=\cos s$
III	$2u=s-\pi /4$ (Euler writes $u=s-45^{\circ }$ then doubles)	$2u=\cos 2u$
IV	$u=s-\pi /2$	$u=\cos u$

So a single transcendental constant

$$s_0=0.7390851332\dots$$

(known to modern readers as the Dottie number) anchors three of the nine chapter-22 problems. Euler computes it once and reuses; the saving is significant in a hand-table era.

Numerical aside

Modern: $s_0$ is the unique real fixed point of $\cos$, transcendental (Lindemann via $\cos s_0=s_0$ algebraic ⇒ $s_0$ algebraic ⇒ $e^{i{s}_0}$ algebraic), and the attractor of the iteration $s_{n+1}=\cos s_n$. Euler does not iterate $\cos$; he uses false position, which converges far faster on a sub-degree window.

Related pages

• chapter-22-on-the-solution-to-several-problems-pertaining-to-the-circle — context.
• false-position-method — the algorithm.
• equal-area-bisection-problems — Problems II, III, IV; III and IV inherit $s_0$.