Arc Equals Versine Plus Sine

Summary: Problem VI of chapter 22 (§536, figure 116). In semicircle $AEB$ find arc $AE$ equal to the sum of versine $AD$ and sine $DE$. Algebraic reduction via the half-angle product identity $1+\cos s+\sin s=2\sqrt{2}\cos \frac{s}{2}\cos (45^{\circ }-\frac{s}{2})$ tames the equation; false-position-method gives $AE=138^{\circ }11^{\prime }53^{\prime \prime }$.

Sources: chapter22, §536. Figure 116 (figures115-118 p. 493).

Last updated: 2026-05-12.

---

Setup

Semicircle $AEB$, center $C$, radius 1. From a point $E$ on the upper arc drop the perpendicular $ED$ to the diameter $AB$. Then

$$DE=\sin s,AD=1-\cos s\text{(versine)},CD=\cos s,$$

where $s=\angle ACE$. We want

$$\text{arc }AE=AD+DE.$$

Euler observes that the arc $AE$ must exceed a quarter circle (otherwise $AE<1+\sin s\le AD+DE$ already plus more). So he reparametrizes by the supplement: let $s=\angle BCE$ (arc $BE$), so arc $AE=180^{\circ }-s$. With $E$ in the second quadrant relative to $A$, $AD=1+\cos s$ and $DE=\sin s$ still, and the equation becomes

$$180^{\circ }-s=1+\cos s+\sin s.$$

Trigonometric reduction

The right-hand side admits a closed product form. Using

$$\sin s=2\sin \frac{s}{2}\cos \frac{s}{2},1+\cos s=2{\cos }^2\frac{s}{2},$$

we factor

$$1+\cos s+\sin s=2\cos \frac{s}{2}(\sin \frac{s}{2}+\cos \frac{s}{2}).$$

The inner sum is itself a single cosine: $\sin \theta +\cos \theta =\sqrt{2}\cos (45^{\circ }-\theta )$, since $\cos (45^{\circ }-\theta )=(\cos \theta +\sin \theta )/\sqrt{2}$. Hence

$$\boxed{180^{\circ }-s=2\sqrt{2}\cos \frac{s}{2}\cos (45^{\circ }-\frac{s}{2}).}$$

This is the form Euler attacks. Its virtue is that every operand is now a single cosine, whose logarithm is directly tabulated.

False position

Try $\frac{s}{2}=20^{\circ }$ and $\frac{s}{2}=21^{\circ }$. Then $45^{\circ }-\frac{s}{2}$ is $25^{\circ }$ and $24^{\circ }$, $180^{\circ }-s$ is $140^{\circ }$ and $138^{\circ }$. Computing LHS - RHS in log form:

$$\begin{array}{rcc}\frac{s}{2}=& 20^{\circ }& 21^{\circ }\\ \log (180^{\circ }-s{)}_{\text{arc}}=& 0.3880054& 0.3817565\\ \log \cos \frac{s}{2}=& 9.9729858& 9.9701517\\ \log \cos (45^{\circ }-\frac{s}{2})=& 9.9572757& 9.9607302\\ \log 2\sqrt{2}=& 0.4515450& 0.4515450\\ \text{cline}2-3\log \text{RHS}=& 0.3818065& 0.3824269\\ \text{error}& +0.0061989& -0.0006704\end{array}$$

Bracket $20^{\circ }<\frac{s}{2}<21^{\circ }$; proportion $68693:61989=1^{\circ }:54^{\prime }$, so $\frac{s}{2}\in (20^{\circ }54^{\prime },20^{\circ }55^{\prime })$.

Minutes pass at $\frac{s}{2}=20^{\circ }54^{\prime }$ and $\frac{s}{2}=20^{\circ }55^{\prime }$ gives errors $+0.0000066$ and $-0.0001065$, proportion $1131:66=1^{\prime }:3^{\prime \prime },30^{\prime \prime \prime }$. So

$$\frac{s}{2}=20^{\circ }54^{\prime }{3}^{\prime \prime }30^{\prime \prime \prime },s=BE=41^{\circ }48^{\prime }{7}^{\prime \prime }{0}^{\prime \prime \prime },AE=138^{\circ }11^{\prime }53^{\prime \prime }{0}^{\prime \prime \prime }.$$

The component lines: $DE=0.6665578$, $AD=1.7454535$. Their sum $2.4120113$ matches the arc length $138.198{\dots }^{\circ }$ in arc units, i.e., $\pi (138.198/180)=2.4120\dots$. ✓

Why this matters for §540

The numerical sine $DE=0.6665578$ in this problem is very close to $\frac{2}{3}=0.6666\overline{6}$, but not equal. Euler points out (§540, see quadrature-commensurability-remark) that if $DE$ had come out to $\frac{2}{3}$ exactly, then $AE=1+\frac{2}{3}+\sqrt{5/9}$ would be algebraic, providing an algebraic length for an arc — i.e., a partial quadrature. Problem VI is the chapter’s poster child for this near-miss.

Figures

Figures 115–118

Related pages

• chapter-22-on-the-solution-to-several-problems-pertaining-to-the-circle — context.
• false-position-method — algorithm.
• quadrature-commensurability-remark — Euler’s §540 reflection turns on this problem.