Machin-like formulas   are useful for efficiently computing numerical approximations for 



π


{\displaystyle \pi }





Task

Verify the following Machin-like formulas are correct by calculating the value of tan   (right hand side) for each equation using exact arithmetic and showing they equal 1:

 





π
4


=
arctan
⁡


1
2


+
arctan
⁡


1
3




{\displaystyle {\pi \over 4}=\arctan {1 \over 2}+\arctan {1 \over 3}}

 
 





π
4


=
2
arctan
⁡


1
3


+
arctan
⁡


1
7




{\displaystyle {\pi \over 4}=2\arctan {1 \over 3}+\arctan {1 \over 7}}


 





π
4


=
4
arctan
⁡


1
5


−
arctan
⁡


1
239




{\displaystyle {\pi \over 4}=4\arctan {1 \over 5}-\arctan {1 \over 239}}


 





π
4


=
5
arctan
⁡


1
7


+
2
arctan
⁡


3
79




{\displaystyle {\pi \over 4}=5\arctan {1 \over 7}+2\arctan {3 \over 79}}


 





π
4


=
5
arctan
⁡


29
278


+
7
arctan
⁡


3
79




{\displaystyle {\pi \over 4}=5\arctan {29 \over 278}+7\arctan {3 \over 79}}


 





π
4


=
arctan
⁡


1
2


+
arctan
⁡


1
5


+
arctan
⁡


1
8




{\displaystyle {\pi \over 4}=\arctan {1 \over 2}+\arctan {1 \over 5}+\arctan {1 \over 8}}

 
 





π
4


=
4
arctan
⁡


1
5


−
arctan
⁡


1
70


+
arctan
⁡


1
99




{\displaystyle {\pi \over 4}=4\arctan {1 \over 5}-\arctan {1 \over 70}+\arctan {1 \over 99}}

 
 





π
4


=
5
arctan
⁡


1
7


+
4
arctan
⁡


1
53


+
2
arctan
⁡


1
4443




{\displaystyle {\pi \over 4}=5\arctan {1 \over 7}+4\arctan {1 \over 53}+2\arctan {1 \over 4443}}


 





π
4


=
6
arctan
⁡


1
8


+
2
arctan
⁡


1
57


+
arctan
⁡


1
239




{\displaystyle {\pi \over 4}=6\arctan {1 \over 8}+2\arctan {1 \over 57}+\arctan {1 \over 239}}


 





π
4


=
8
arctan
⁡


1
10


−
arctan
⁡


1
239


−
4
arctan
⁡


1
515




{\displaystyle {\pi \over 4}=8\arctan {1 \over 10}-\arctan {1 \over 239}-4\arctan {1 \over 515}}


 





π
4


=
12
arctan
⁡


1
18


+
8
arctan
⁡


1
57


−
5
arctan
⁡


1
239




{\displaystyle {\pi \over 4}=12\arctan {1 \over 18}+8\arctan {1 \over 57}-5\arctan {1 \over 239}}


 





π
4


=
16
arctan
⁡


1
21


+
3
arctan
⁡


1
239


+
4
arctan
⁡


3
1042




{\displaystyle {\pi \over 4}=16\arctan {1 \over 21}+3\arctan {1 \over 239}+4\arctan {3 \over 1042}}


 





π
4


=
22
arctan
⁡


1
28


+
2
arctan
⁡


1
443


−
5
arctan
⁡


1
1393


−
10
arctan
⁡


1
11018




{\displaystyle {\pi \over 4}=22\arctan {1 \over 28}+2\arctan {1 \over 443}-5\arctan {1 \over 1393}-10\arctan {1 \over 11018}}


 





π
4


=
22
arctan
⁡


1
38


+
17
arctan
⁡


7
601


+
10
arctan
⁡


7
8149




{\displaystyle {\pi \over 4}=22\arctan {1 \over 38}+17\arctan {7 \over 601}+10\arctan {7 \over 8149}}


 





π
4


=
44
arctan
⁡


1
57


+
7
arctan
⁡


1
239


−
12
arctan
⁡


1
682


+
24
arctan
⁡


1
12943




{\displaystyle {\pi \over 4}=44\arctan {1 \over 57}+7\arctan {1 \over 239}-12\arctan {1 \over 682}+24\arctan {1 \over 12943}}


 





π
4


=
88
arctan
⁡


1
172


+
51
arctan
⁡


1
239


+
32
arctan
⁡


1
682


+
44
arctan
⁡


1
5357


+
68
arctan
⁡


1
12943




{\displaystyle {\pi \over 4}=88\arctan {1 \over 172}+51\arctan {1 \over 239}+32\arctan {1 \over 682}+44\arctan {1 \over 5357}+68\arctan {1 \over 12943}}


and confirm that the following formula is incorrect by showing   tan   (right hand side)   is not   1:

 





π
4


=
88
arctan
⁡


1
172


+
51
arctan
⁡


1
239


+
32
arctan
⁡


1
682


+
44
arctan
⁡


1
5357


+
68
arctan
⁡


1
12944




{\displaystyle {\pi \over 4}=88\arctan {1 \over 172}+51\arctan {1 \over 239}+32\arctan {1 \over 682}+44\arctan {1 \over 5357}+68\arctan {1 \over 12944}}


These identities are useful in calculating the values:

 



tan
⁡
(
a
+
b
)
=



tan
⁡
(
a
)
+
tan
⁡
(
b
)


1
−
tan
⁡
(
a
)
tan
⁡
(
b
)





{\displaystyle \tan(a+b)={\tan(a)+\tan(b) \over 1-\tan(a)\tan(b)}}


 



tan
⁡

(
arctan
⁡


a
b


)

=


a
b




{\displaystyle \tan \left(\arctan {a \over b}\right)={a \over b}}


 



tan
⁡
(
−
a
)
=
−
tan
⁡
(
a
)


{\displaystyle \tan(-a)=-\tan(a)}




You can store the equations in any convenient data structure, but for extra credit parse them from human-readable text input.

Note: to formally prove the formula correct, it would have to be shown that 






−
3
p
i

4




{\displaystyle {-3pi \over 4}}

 < right hand side < 






5
p
i

4




{\displaystyle {5pi \over 4}}

 due to 



tan
⁡
(
)


{\displaystyle \tan()}

 periodicity.



(ns tanevaulator
  (:gen-class))
 
;; Notation: [a b c] -> a x arctan(a/b)
(def test-cases   [
                          [[1, 1, 2], [1, 1, 3]],
                          [[2, 1, 3], [1, 1, 7]],
                          [[4, 1, 5], [-1, 1, 239]],
                          [[5, 1, 7], [2, 3, 79]],
                          [[1, 1, 2], [1, 1, 5], [1, 1, 8]],
                          [[4, 1, 5], [-1, 1, 70], [1, 1, 99]],
                          [[5, 1, 7], [4, 1, 53], [2, 1, 4443]],
                          [[6, 1, 8], [2, 1, 57], [1, 1, 239]],
                          [[8, 1, 10], [-1, 1, 239], [-4, 1, 515]],
                          [[12, 1, 18], [8, 1, 57], [-5, 1, 239]],
                          [[16, 1, 21], [3, 1, 239], [4, 3, 1042]],
                          [[22, 1, 28], [2, 1, 443], [-5, 1, 1393], [-10, 1, 11018]],
                          [[22, 1, 38], [17, 7, 601], [10, 7, 8149]],
                          [[44, 1, 57], [7, 1, 239], [-12, 1, 682], [24, 1, 12943]],
                          [[88, 1, 172], [51, 1, 239], [32, 1, 682], [44, 1, 5357], [68, 1, 12943]],
                          [[88, 1, 172], [51, 1, 239], [32, 1, 682], [44, 1, 5357], [68, 1, 12944]]
                  ])
 
(defn tan-sum [a b]
  " tan (a + b) "
  (/ (+ a b) (- 1 (* a b))))
 
(defn tan-eval [m]
  " Evaluates tan of a triplet (e.g. [1, 1, 2])"
  (let [coef (first m)
        rat (/ (nth m 1) (nth m 2))]
  (cond
    (= 1  coef) rat
    (neg? coef) (tan-eval [(- (nth m 0)) (- (nth m 1)) (nth m 2)])
    :else (let [
                ca (quot coef 2)
                cb (- coef ca)
                a (tan-eval [ca (nth m 1) (nth m 2)])
                b (tan-eval [cb (nth m 1) (nth m 2)])]
            (tan-sum a b)))))
 
(defn tans [m]
  " Evaluates tan of set of triplets (e.g. [[1, 1, 2], [1, 1, 3]])"
  (if (= 1 (count m))
    (tan-eval (nth m 0))
    (let [a (tan-eval (first m))
          b (tans (rest m))]
      (tan-sum a b))))
 
(doseq [q test-cases]
  " Display results "
  (println "tan " q " = "(tans q)))
 