数学 - Python - 円の中にひそむ関数 - 三角関数 - 加法定理 - 三角関数の諸公式 - 正弦、余弦、正接、円周率、倍角、半角、和を積に変形

学習環境

新装版数学読本2 (松坂和夫(著)、岩波書店)の第8章(円の中にひそむ関数 - 三角関数)、8.2(加法定理)、三角関数の諸公式の問37の解答を求めてみる。

1. $\begin{array}{l} \cos 2 A + \cos 2 B + \cos^{2} C \\ = \cos ((A + B) + (A - B)) + \cos ((A + B) - (A - B)) \\ + \cos (2 π - 2 (A + B)) \\ = 2 \cos (A + B) \cos (A - B) + \cos 2 (A + B) \\ = 2 \cos (A + B) \cos (A - B) + (2 \cos^{2} (A + B) - 1) \\ = - 1 + 2 \cos (A + B) (\cos (A - B) + \cos (A + B)) \\ = - 1 + 2 \cos (π - C) \cos A \cos B \\ = - 1 + 2 \cos A \cos B \cos C \end{array}$
2. $\begin{array}{l} \sin A + \sin B + \sin C \\ = \sin ((\frac{A}{2} + \frac{B}{2}) + (\frac{A}{2} - \frac{B}{2})) + \sin ((\frac{A}{2} + \frac{B}{2}) - (\frac{A}{2} - \frac{B}{2})) \\ + \sin (π - (A + B)) \\ = 2 \sin (\frac{A}{2} + \frac{B}{2}) \cos (\frac{A}{2} - \frac{B}{2}) + \sin (A + B) \\ = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2} + \sin (2 \cdot \frac{A + B}{2}) \\ = 2 \sin \frac{A + B}{2} \cos (\frac{A}{2} - \frac{B}{2}) + 2 \sin \frac{A + B}{2} \cos \frac{A + B}{2} \\ = 2 \frac{A + B}{2} (\cos \frac{A - B}{2} + \cos \frac{A + B}{2}) \\ = 2 (\sin \frac{π - C}{2}) (\cos \frac{A - B}{2} + \cos \frac{A + B}{2}) \\ = 2 \cos \frac{C}{2} (\cos \frac{A - B}{2} + \cos \frac{A + B}{2}) \\ = 2 \cos (\frac{C}{2}) \cdot 2 \cos \frac{A}{2} \cos \frac{B}{2} \\ = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \end{array}$
3. $\begin{array}{l} \cos A + \cos B + \cos C \\ = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2} + \cos (π - (A + B)) \\ = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2} - \cos (A + B) \\ = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2} - \cos 2 \cdot \frac{A + B}{2} \\ = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2} - (\cos^{2} \frac{A + B}{2} - \sin^{2} \frac{A + B}{2}) \\ = 2 \cos \frac{A + B}{2} \cos \frac{A - B}{2} - (2 \cos^{2} \frac{A + B}{2} - 1) \\ = 1 + 2 \cos \frac{A + B}{2} (\cos \frac{A - B}{2} - \cos \frac{A + B}{2}) \\ = 1 + 2 \cos \frac{π - C}{2} \cdot 2 \sin \frac{A}{2} \sin \frac{B}{2} \\ = 1 + 4 \sin \frac{A}{2} \sin \frac{B}{2} (- \sin (- \frac{C}{2})) \\ = 1 + 4 c i n \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \end{array}$
4. $\begin{array}{l} \tan A + \tan B + \tan C \\ = \tan A + \tan B + \tan (π - (A + B)) \\ = \tan A + \tan B - \tan (A + B) \\ = \tan A + \tan B - \frac{\tan A + \tan B}{1 - \tan A \tan B} \\ = \frac{(- \tan A \tan B) (\tan A + \tan B)}{1 - \tan A \tan B} \\ = - \tan A \tan B \tan (A + B) \\ = - \tan A \tan B \tan (π - C) \\ = - \tan A \tan B (- \tan C) \\ = \tan A \tan B \tan C \end{array}$
  （証明終）

#!/usr/bin/env python3 from unittest import TestCase, main from sympy import sin, cos, tan, Rational, pi print('37.') a = 1 b = 2 c = pi - (a + b) class MyTestCase(TestCase): def test1(self): self.assertEqual(float(cos(2 * a) + cos(2 * b) + cos(2 * c)), float(-1 - 4 * cos(a) * cos(b) * cos(c))) def test2(self): self.assertAlmostEqual(float(sin(a) + sin(b) + sin(c)), float(4 * cos(a / 2) * cos(b / 2) * cos(c / 2))) def test3(self): self.assertEqual(float(cos(a) + cos(b) + cos(c)), float(1 + 4 * sin(a / 2) * sin(b / 2) * sin(c / 2))) def test4(self): self.assertEqual(float(tan(a) + tan(b) + tan(c)), float(tan(a) * tan(b) * tan(c))) if __name__ == '__main__': main()

入出力結果(Zsh、cmd.exe(コマンドプロンプト)、Terminal、Jupyter(IPython))

% ./sample37.py -v 37. test1 (__main__.MyTestCase) ... ok test2 (__main__.MyTestCase) ... ok test3 (__main__.MyTestCase) ... ok test4 (__main__.MyTestCase) ... ok ---------------------------------------------------------------------- Ran 4 tests in 0.029s OK %

Kamimura's blog

2019年12月10日火曜日

数学 - Python - 円の中にひそむ関数 - 三角関数 - 加法定理 - 三角関数の諸公式 - 正弦、余弦、正接、円周率、倍角、半角、和を積に変形

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