第14章§6 - 4

（証明） 問題1で示した$\tan$の加法定理より， $$\begin{array}{rl}\tan 2x& =\tan (x+x)=\frac{2\tan x}{1-{\tan }^{2}x}\\ \tan 4x& =\tan (2x+2x)=\frac{2\tan 2x}{1-{\tan }^{2}2x}\\ & =\frac{2\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)}{1-{\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)}^2}\\ & =\frac{4\tan x-4{\tan }^{3}x}{1-6{\tan }^{2}x+{\tan }^{4}x}\end{array}$$ よって $$\begin{array}{r}\tan (4\arctan \frac{1}{5})=\frac{4\cdot \frac{1}{5}-4{\left(\frac{1}{5}\right)}^3}{1-6{\left(\frac{1}{5}\right)}^2+{\left(\frac{1}{5}\right)}^4}=\frac{120}{119}\end{array}$$ となるから $$\begin{array}{rl}& \tan (4\arctan \frac{1}{5}-\arctan \frac{1}{239})\\ & =\frac{\tan (4\arctan \frac{1}{5})-\tan (\arctan \frac{1}{239})}{1-\tan (4\arctan \frac{1}{5})\tan (\arctan \frac{1}{239})}\\ & =\frac{\frac{120}{119}-\frac{1}{239}}{1-\frac{120}{119}\frac{1}{239}}=\frac{120\cdot 239-119}{119\cdot 239+120}=\frac{28561}{28561}\\ & =1\end{array}$$ ゆえに $$\begin{array}{c}\arctan 1=4\arctan \frac{1}{5}-\arctan \frac{1}{239}\\ \therefore \frac{\pi }{4}=4\arctan \frac{1}{5}-\arctan \frac{1}{239}\end{array}$$ （証明終）