Sorry, no pictures. Explanation here.
We wish to trisect the given angle $AOB$. Assume the angle is less than 90 degrees and $OA=OB$; see the diagram above.
Let $\alpha$ and $\beta$ be the sizes of the angles $AOB$ and $FOT$, respectively. It is straightforward to show that \[ \beta = \frac{\sin \alpha}{2+\cos \alpha} = \frac{\alpha}{3} - \frac{1}{2^2 \cdot 3^3 \cdot 5} \alpha^5 + O(\alpha^7) = \frac{\alpha}{3} - \frac{1}{540} \alpha^5 + O(\alpha^7). \] The error $ \ds e(\alpha) = \frac{\alpha}{3} - \beta $ is monotonically increasing in $\alpha$. The worst error on the interval $0 \le \alpha \le \pi/2$ is $e(\pi/2) =$ 0.0236 radians = 1.352 degrees.
It is interesting that the error is $O(\alpha^5)$ rather than $O(\alpha^3)$ as one might have expected. Despite this, the method's accuracy is not particularly remarkable for angles that are not very close to zero.
This applet was created by
Rouben Rostamian
using
David Joyce's
Geometry
Applet
on July 26, 2002.
Cosmetic revisions on June 6, 2010.
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