数学 - Python - 円の中にひそむ関数 - 三角関数 - 加法定理 - 正弦・余弦の加法定理 - 弧度法、radian

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)、正弦・余弦の加法定理の問16の解答を求めてみる。

1. $\begin{array}{l} \sin 1 5^{\circ} \\ = \sin (6 0^{\circ} - 4 5^{\circ}) \\ = \sin (\frac{1}{3} π - \frac{1}{4} π) \\ = \sin \frac{π}{3} \cos \frac{π}{4} - \cos \frac{π}{3} \sin \frac{π}{4} \\ = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} - \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \\ = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\ = \frac{\sqrt{6} - \sqrt{2}}{4} \end{array}$
2. $\begin{array}{l} \cos 1 5^{\circ} \\ = \cos \frac{π}{12} \\ = \cos (\frac{1}{3} π - \frac{1}{4} π) \\ = \cos \frac{π}{3} \cos \frac{π}{4} + \sin \frac{π}{3} \sin \frac{π}{4} \\ = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} \\ = \frac{\sqrt{3} + 1}{2 \sqrt{2}} \\ = \frac{\sqrt{6} + \sqrt{2}}{4} \end{array}$
3. $\begin{array}{l} \sin 10 5^{\circ} \\ = \sin \frac{7}{12} π \\ = \sin (\frac{π}{2} + \frac{π}{12}) \\ = \sin \frac{π}{2} \cos \frac{π}{12} + \cos \frac{π}{2} \sin \frac{π}{12} \\ = \frac{\sqrt{6} + \sqrt{2}}{4} \end{array}$
4. $\begin{array}{l} \cos 10 5^{\circ} \\ = \cos \frac{7}{12} π \\ = - \sin (\frac{7}{12} π - \frac{π}{2}) \\ = - \sin \frac{π}{12} \\ = \frac{\sqrt{2} - \sqrt{6}}{4} \end{array}$
5. $\begin{array}{l} \sin 12 0^{\circ} \\ = \sin \frac{2}{3} π \\ = \frac{\sqrt{3}}{2} \end{array}$
6. $\begin{array}{l} \cos 12 0^{\circ} \\ = \cos \frac{2}{3} π \\ = - \frac{1}{2} \end{array}$
7. $\begin{array}{l} \sin 16 5^{\circ} \\ = \sin \frac{165}{180} π \\ = \sin \frac{11}{12} π \\ = \sin (\frac{10}{12} π + \frac{1}{12} π) \\ = \sin (\frac{5}{6} π + \frac{π}{12}) \\ = \sin \frac{5}{6} π \cos \frac{π}{12} + \cos \frac{5}{6} π \sin \frac{π}{12} \\ = \frac{1}{2} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{6} - \sqrt{2}}{4} \\ = \frac{\sqrt{6} + \sqrt{2} - \sqrt{18} + \sqrt{6}}{8} \\ = \frac{2 \sqrt{6} + \sqrt{2} - 3 \sqrt{2}}{8} \\ = \frac{2 \sqrt{6} - 2 \sqrt{2}}{8} \\ = \frac{\sqrt{6} - \sqrt{2}}{4} \end{array}$
8. $\begin{array}{l} \cos 16 5^{\circ} \\ = \cos \frac{11}{12} π \\ = - \cos (\frac{11}{12} π - π) \\ = - \cos (- \frac{π}{12}) \\ = - \cos \frac{π}{12} \\ = - \frac{\sqrt{6} + \sqrt{2}}{4} \end{array}$

#!/usr/bin/env python3 import math from unittest import TestCase, main from sympy import pprint, symbols, sin, cos, sqrt, Rational print('16.') class MyTestCase(TestCase): def test(self): angles = [15, 105, 120, 165] fs = [sin, cos] results = [(sqrt(6) - sqrt(2)) / 4, (sqrt(6) + sqrt(2)) / 4, (sqrt(6) + sqrt(2)) / 4, (sqrt(2) - sqrt(6)) / 4, sqrt(3) / 2, -Rational(1, 2), (sqrt(6) - sqrt(2)) / 4, -(sqrt(6) + sqrt(2)) / 4] for i, angle in enumerate(angles): for j, f in enumerate(fs): self.assertAlmostEqual(f(math.radians(angle)), float(results[2 * i + j])) if __name__ == '__main__': main()

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

Kamimura's blog

2019年11月9日土曜日

数学 - Python - 円の中にひそむ関数 - 三角関数 - 加法定理 - 正弦・余弦の加法定理 - 弧度法、radian

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