Proving $\pi = 48\tan^{-1}\frac{1}{18} + 32 \tan^{-1}\frac{1}{57} - 20\tan^{-1}\frac{1}{239}$

Question

The below equation represents $\pi$ to some decimals using tangent inverse. I need to prove that the left hand side of the equation equals the right hand side. $$ \pi = 48\tan^{-1}\frac{1}{18} + 32 \tan^{-1}\frac{1}{57} - 20\tan^{-1}\frac{1}{239} $$

Answer 1

To prove given formula, we'll use identity (see here) $$ \arctan\frac{a_1}{b_1} + \arctan\frac{a_2}{b_2} = \arctan\frac{a_1 b_2+a_2b_1}{b_1b_2-a_1a_2}. \tag{1} $$ Denote $$\arctan\frac{1}{b} = f(b).\tag{2}$$ Then given formula will have form $$ f(1) = 12 f(18) + 8 f(57) - 5 f(239).\tag{3} $$

The proof will be based on $2$-term-RHS identities which can be easily checked via $(1)$.

a): since $f(239)=4f(5)-f(1)$ (see Machin's Formula), after substituting $f(239)$ to $(3)$ we get equivalent formula to prove: $$ f(1) = 5f(5)-3f(18)-2f(57).\tag{4} $$

b): since $f(18)=f(5)-f(7)$, we get new identity to prove: $$ f(1)=2f(5)+3f(7)-2f(57).\tag{5} $$

c): since $f(57)=f(7)-f(8)$, we came to new identity to prove: $$ f(1) = 2f(5)+f(7)+2f(8).\tag{6} $$

d): since $f(8)=f(3)-f(5)$, we came to $2$-term-RHS formula: $$ f(1)=2f(3)+f(7),\tag{7} $$

e): and since $f(7)=f(2)-f(3)$, we came to another well-known Machin-like (Euler's) formula: $$ f(1)=f(2)+f(3).\tag{8} $$

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So, given formula can be obtained from $(8)$: each step replaces argument to more appropriate two arguments (and replace $5$ at the final step). The sketch is:

$(2),3 \longrightarrow \color{red}{3},\color{red}{7}$

$(3),7 \longrightarrow \color{red}{5},7,\color{red}{8}$

$5,7,(8) \longrightarrow 5,\color{red}{7},\color{red}{57}$

$5,(7),57 \longrightarrow \color{red}{5},\color{red}{18},57$

$(5),18,57 \longrightarrow \color{red}{1},18,57,\color{red}{239}$.

Answer 2

By brute force:

Using the addition formula ($\tan(\arctan a+\arctan b)=\dfrac{a+b}{1-ab}$),

$$\tan\left(12\arctan\frac1{18}\right)=\tan\left(6\arctan\frac{36}{323}\right)=\tan\left(3\arctan\frac{23256}{103033}\right) \\=\tan\left(\arctan\frac{23256}{103033}+2\arctan\frac{23256}{103033}\right) \\=\tan\left(\arctan\frac{23256}{103033}+\arctan\frac{4792270896}{10074957553}\right)=\frac{728065260080136}{926604049600873},$$

$$\tan\left(8\arctan\frac1{57}\right)=\tan\left(4\arctan\frac{57}{1624}\right)=\tan\left(2\arctan\frac{185136}{2634127}\right)=\frac{975343472544}{6904349713633},$$

$$\tan\left(\arctan\frac{728065260080136}{926604049600873}+\arctan\frac{975343472544}{6904349713633}\right)=\frac{99498527400}{95420159401},$$

$$\tan\left(4\arctan\frac1{239}\right)=\tan\left(2\arctan\frac{239}{28260}\right)=\frac{13651680}{815616479},$$

$$\tan\left(\arctan\frac1{239}+\arctan\frac{13651680}{815616479}\right)=\frac{4078367999}{194918686801},$$

and finally

$$\tan\left(\arctan\frac{99498527400}{95420159401}-\arctan\frac{4078367999}{194918686801}\right)=1.$$

Easy as that.