近似三等分任意角的尺规作图

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设∠AOB为已知角，以O为圆心画圆弧交两边于A，B，BO的延长线交圆于D.作∠AOB的平分线OC，取OD的中点E，连结EC，则$\angle CED≈\frac13 \angle AOB$，当0<∠AOB≤90°时，绝对误差：$\angle CED-\frac{\angle AOB}3 \in (0,21'40.296'')$，此函数递增，∠AOB→0时左边取等，∠AOB→90°时右边取等，精确值为${\tan ^{ - 1}}\left( {2 - \sqrt 2 } \right) - \frac{\pi }{6}$，相对误差：$\frac{\angle CED-\frac{\angle AOB}3}{\frac{\angle AOB}3} \in (0,0.0120398)$，此函数递增，∠AOB→0时左边取等，∠AOB→90°时右边取等，精确值为$\frac{{6{{\tan }^{ - 1}}(2 - \sqrt 2 )}}{\pi } - 1$

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murderousmaths上的作法

How to trisect an angle without cheating? Sent in by Joe Maddox, Ocala, Florida, USA
1/ We can start with any angle in between about 30º and 90º. Here we've labelled the angle ABC. Using compasses measure off three equal sections along BC labelled 1,2,3.	2/ Draw two big arcs centred on B, one to pass through point 2 and the other to pass point 3. You now bisect the angle ABC. One way to do it is to draw in two lines between the points where the arcs cross the lines AB and BC, then draw the bisector BD through the point where these new lines cross. All nice and easy so far.
3/ Now you bisect angle ABD. You can use cross lines again, and you end up drawing line BE. After that, you draw a circle. The centre point is X which is where BE crosses the "2" arc, and the circle goes through point B. (So the circle will have the same radius as the 2 arc.) The circle cuts BD at point Y.	4/ You now draw a line through Y parallel to BC. (If you don't know how to construct parallel lines with a straight edge and compasses, you'll find it in Savage Shapes. Sorry, we don't have space to tell you here!) This parallel line cuts the 3 arc at point Z.
5/ You draw a line from B that goes through Z to P. This line divides the angle in the ratio 1:2 !	6/ Finally you bisect the angle PBC and draw line BQ. The angle has been trisected perfectly ... or has it?
If you draw an angle then follow the instructions, this method seems to work perfectly - even though it has been proved beyond any doubt that trisecting an angle only using compasses and a straight edge is impossible! So what's wrong? We tested out Joe's method using trigonometry. (If you've read The Fiendish Angletron you'll know all about this stuff.) The most important part is when the line BP appears. The angle PBC should be exactly 2/3 of the angle ABC. We worked out a formula to show what angle this method produces. Here we've called the size of angle ABC = J and PBC = T. Next we chose some angles and tried them out in the formula. Here are the results! Angle J What angle T should be What angle T really is! 90º 60º 60.580º 60º 40º 40.087º 24º 16º 16.004º This means the error when J=90 is less than 1%, and for smaller angles it's even less! So Joe's method is really really close but sadly it's not exact. Never mind - Joe tells us he's working on an even better version!

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$$\sin^{-1}\left\{\frac43\cos\frac x4\sin\frac x2\right\}=\frac{2 x}3 + \frac{x^3}{1296} + \frac{11 x^5}{27648} +\mathcal O(x^7)
\quad\text{(Taylor series)}$$