Quantcast
Channel: 円周率近似値の日に生まれて理系じゃないわけないだろ! - knifeのblog
Viewing all articles
Browse latest Browse all 5376

18度系と15度系から加法定理で3度系を網羅する

$
0
0

15度系から5度系、18度系から6度系を、それぞれ1/3倍角でカルダノ解法で求めたが、
3乗根が出てくるので、その後の展開にこまる。

というわけで、タイトルの18度系と15度系を使って加法定理を解くと、3度系を網羅出来るという算段です。

 

cos(0˚)=
sin(90˚)=
+1
cos(3˚)=
cos(357˚)=
sin(93˚)=
sin(87˚)=
+
(6+2)10+25+(6-2)(-1+5)
16
cos(6˚)=
cos(354˚)=
sin(96˚)=
sin(84˚)=
+
15+3+10-25
8
cos(9˚)=
cos(351˚)=
sin(99˚)=
sin(81˚)=
+
10+2+25-5
8
cos(12˚)=
cos(348˚)=
sin(102˚)=
sin(78˚)=
+
-1+5+30+65
8
cos(15˚)=
cos(345˚)=
sin(105˚)=
sin(75˚)=
+
6+2
4
cos(18˚)=
cos(342˚)=
sin(108˚)=
sin(72˚)=
+
10+25
4
cos(21˚)=
cos(339˚)=
sin(111˚)=
sin(69˚)=
+
(6+2)(1+5)+(6-2)10-25
16
cos(24˚)=
cos(336˚)=
sin(114˚)=
sin(66˚)=
+
1+5+30-65
8
cos(27˚)=
cos(333˚)=
sin(117˚)=
sin(63˚)=
+
10-2+25+5
8
cos(30˚)=
cos(330˚)=
sin(120˚)=
sin(60˚)=
+
3
2
cos(33˚)=
cos(327˚)=
sin(123˚)=
sin(57˚)=
+
(6+2)10+25-(6-2)(-1+5)
16
cos(36˚)=
cos(324˚)=
sin(126˚)=
sin(54˚)=
+
1+5
4
cos(39˚)=
cos(321˚)=
sin(129˚)=
sin(51˚)=
+
(6+2)10-25+(6-2)(1+5)
16
cos(42˚)=
cos(318˚)=
sin(132˚)=
sin(48˚)=
+
15-3+10+25
8
cos(45˚)=
cos(315˚)=
sin(135˚)=
sin(45˚)=
+
2
2
cos(48˚)=
cos(312˚)=
sin(138˚)=
sin(42˚)=
+
1-5+30+65
8
cos(51˚)=
cos(309˚)=
sin(141˚)=
sin(39˚)=
+
(6+2)(1+5)-(6-2)10-25
16
cos(54˚)=
cos(306˚)=
sin(144˚)=
sin(36˚)=
+
10-25
4
cos(57˚)=
cos(303˚)=
sin(147˚)=
sin(33˚)=
+
(6+2)(-1+5)+(6-2)10+25
16
cos(60˚)=
cos(300˚)=
sin(150˚)=
sin(30˚)=
+
1
2
cos(63˚)=
cos(297˚)=
sin(153˚)=
sin(27˚)=
+
-10+2+25+5
8
cos(66˚)=
cos(294˚)=
sin(156˚)=
sin(24˚)=
+
15+3-10-25
8
cos(69˚)=
cos(291˚)=
sin(159˚)=
sin(21˚)=
+
(6+2)10-25-(6-2)(1+5)
16
cos(72˚)=
cos(288˚)=
sin(162˚)=
sin(18˚)=
+
-1+5
2
cos(75˚)=
cos(285˚)=
sin(165˚)=
sin(15˚)=
+
6-2
4
cos(78˚)=
cos(282˚)=
sin(168˚)=
sin(12˚)=
+
-15+3+10+25
8
cos(81˚)=
cos(279˚)=
sin(171˚)=
sin(9˚)=
+
10+2-25-5
8
cos(84˚)=
cos(276˚)=
sin(174˚)=
sin(6˚)=
+
-1-5+30-65
8
cos(87˚)=
cos(273˚)=
sin(177˚)=
sin(3˚)=
+
(6+2)(-1+5)-(6-2)10+25
16
cos(90˚)=
cos(270˚)
sin(180˚)=
sin(0˚)=
±0
cos(93˚)=
cos(261˚)=
sin(195˚)=
sin(357˚)=
-
(6+2)(-1+5)-(6-2)10+25
16
cos(96˚)=
cos(258˚)=
sin(198˚)=
sin(354˚)=
-
-1-5+30-65
8
cos(99˚)=
cos(261˚)=
sin(195˚)=
sin(351˚)=
-
10+2-25-5
8
cos(102˚)=
cos(258˚)=
sin(192˚)=
sin(348˚)=
-
-15+3+10+25
8
cos(105˚)=
cos(255˚)=
sin(195˚)=
sin(345˚)=
-
6-2
4
cos(108˚)=
cos(252˚)=
sin(198˚)=
sin(342˚)=
-
-1+5
2
cos(111˚)=
cos(243˚)=
sin(201˚)=
sin(339˚)=
-
(6+2)10-25-(6-2)(1+5)
16
cos(114˚)=
cos(246˚)=
sin(204˚)=
sin(336˚)=
-
15+3-10-25
8
cos(117˚)=
cos(243˚)=
sin(207˚)=
sin(333˚)=
-
-10+2+25+5
8
cos(120˚)=
cos(240˚)=
sin(210˚)=
sin(330˚)=
-
1
2
cos(123˚)=
cos(237˚)=
sin(213˚)=
sin(327˚)=
-
(6+2)(-1+5)+(6-2)10+25
16
cos(126˚)=
cos(234˚)=
sin(216˚)=
sin(324˚)=
-
10-25
4
cos(129˚)=
cos(231˚)=
sin(219˚)=
sin(321˚)=
-
(6+2)(1+5)-(6-2)10-25
16
cos(132˚)=
cos(228˚)=
sin(222˚)=
sin(318˚)=
-
1-5+30+65
8
cos(135˚)=
cos(225˚)=
sin(225˚)=
sin(315˚)=
-
2
2
cos(138˚)=
cos(222˚)=
sin(228˚)=
sin(312˚)=
-
15-3+10+25
8
cos(141˚)=
cos(219˚)=
sin(231˚)=
sin(309˚)=
-
(6+2)10-25+(6-2)(1+5)
16
cos(144˚)=
cos(216˚)=
sin(234˚)=
sin(306˚)=
-
1+5
4
cos(147˚)=
cos(213˚)=
sin(237˚)=
sin(303˚)=
-
(6+2)10+25-(6-2)(-1+5)
16
cos(150˚)=
cos(210˚)=
sin(240˚)=
sin(300˚)=
-
3
2
cos(153˚)=
cos(207˚)=
sin(243˚)=
sin(291˚)=
+
10-2+25+5
8
cos(156˚)=
cos(204˚)=
sin(246˚)=
sin(294˚)=
-
1+5+30-65
8
cos(159˚)=
cos(201˚)=
sin(249˚)=
sin(291˚)=
-
(6+2)(1+5)+(6-2)10-25
16
cos(162˚)=
cos(198˚)=
sin(252˚)=
sin(288˚)=
-
10+25
4
cos(165˚)=
cos(195˚)=
sin(255˚)=
sin(285˚)=
-
6+2
4
cos(168˚)=
cos(192˚)=
sin(258˚)=
sin(282˚)=
-
-1+5+30+65
8
cos(171˚)=
cos(189˚)=
sin(261˚)=
sin(279˚)=
-
10+2+25-5
8
cos(174˚)=
cos(186˚)=
sin(264˚)=
sin(276˚)=
-
15+3+10-25
8
cos(177˚)=
cos(183˚)=
sin(267˚)=
sin(273˚)=
-
(6+2)10+25+(6-2)(-1+5)
16
cos(180˚)=
sin(270˚)=
-1


さて、こういう式をユニーク(一意)にするために、法則を設けることとした。
まずは、分母の有理化は行うものとして、
根号や括弧において、降べきの順にならべるのだが、
整数、平方根、二重根号
の順である。
整数同士は加減算出来るために項は1つにまとまるが、平方根は根の中の大小で降順とし、根の符号には左右されませんので注意が必要。
また、共通の平方根の約数を持っていたとしても、根号や括弧の外に出さないこととし、出せるのは整数のみとした。
また、符号は明示的に分数の前に付けたので、分数部分は常に正となるようになっている。

今更、加法定理の説明は必要ないとは思うが、
cos(α±β) = cos(α)cos(β) sin(α)sin(β)
を使って求めた。
別にsinを使っても結果は同じになるだろうが、私の頭の中での混乱や計算ミスが、より少ない方を使ったに過ぎない。

さて、3度系を網羅したから、ここから1/3倍角でカルダノの解法を使って1度系を網羅しますか?
流石に暇を持て余した私でも、そこまではやりませんよ。
もう3乗根はお腹いっぱいですので、


ではでは


Viewing all articles
Browse latest Browse all 5376

Trending Articles