数学 - Python - SymPy - 解析学 - 各種の初等関数 - 三角関数(正弦、余弦、正接、加法定理、半角の公式)

解析入門〈1〉

学習環境

• Surface 3 (4G LTE)、Surface 3 タイプ カバー、Surface ペン(端末)
• Windows 10 Pro (OS)
• 数式入力ソフト(TeX, MathML): MathType
• MathML対応ブラウザ: Firefox、Safari
• MathML非対応ブラウザ(Internet Explorer, Google Chrome...)用JavaScript Library: MathJax
• Pythonからはじめる数学入門(参考書籍)

解析入門〈1〉(松坂 和夫(著)、岩波書店)の第5章(各種の初等関数)、5.3(三角関数)、問題5.3-5.を取り組んでみる。

6. 
   $\begin{array}{l}\sin \frac{\pi }{12}\\ =\sin \left(\frac{\pi }{3}-\frac{\pi }{4}\right)\\ =\sin \frac{\pi }{3}\cos \frac{\pi }{4}-\cos \frac{\pi }{3}\sin \frac{\pi }{4}\\ =\frac{\sqrt{3}}{2}·\frac{1}{\sqrt{2}}-\frac{1}{2}\frac{1}{\sqrt{2}}\\ =\frac{\sqrt{3}-1}{2\sqrt{2}}\\ =\frac{\sqrt{6}-\sqrt{2}}{4}\\ \\ \cos \frac{\pi }{12}\\ =\cos \left(\frac{\pi }{3}-\frac{\pi }{4}\right)\\ =\cos \frac{\pi }{3}\cos \frac{\pi }{4}+\sin \frac{\pi }{3}\sin \frac{\pi }{4}\\ =\frac{1}{2}\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}·\frac{1}{\sqrt{2}}\\ =\frac{\sqrt{6}+\sqrt{2}}{4}\\ \\ \tan \frac{\pi }{12}\\ =\frac{=\frac{\sqrt{6}-\sqrt{2}}{4}}{\frac{\sqrt{6}+\sqrt{2}}{4}}\\ =\frac{6+2-2\sqrt{12}}{4}\\ =2-\sqrt{3}\end{array}$
7. 
   $\begin{array}{l}{\sin }^2\frac{\pi }{8}\\ ={\sin }^2\frac{\frac{\pi }{4}}{2}\\ =\frac{1-\cos \frac{\pi }{4}}{2}\\ =\frac{1-\frac{1}{\sqrt{2}}}{2}\\ =\frac{2-\sqrt{2}}{4}\\ \sin \frac{\pi }{8}=\frac{\sqrt{2-\sqrt{2}}}{2}\\ \\ {\cos }^2\frac{\pi }{8}\\ ={\sin }^2\frac{\frac{\pi }{4}}{2}\\ =\frac{1+\cos \frac{\pi }{4}}{2}\\ =\frac{1+\frac{1}{\sqrt{2}}}{2}\\ =\frac{2+\sqrt{2}}{4}\\ \cos \frac{\pi }{8}=\frac{\sqrt{2+\sqrt{2}}}{2}\\ \\ \tan \frac{\pi }{8}\\ =\frac{\frac{\sqrt{2-\sqrt{2}}}{2}}{\frac{\sqrt{2+\sqrt{2}}}{2}}\\ =\frac{\sqrt{2}}{2+\sqrt{2}}\\ =\frac{2\sqrt{2}-2}{4-2}\\ =\sqrt{2}-1\end{array}$

コード(Emacs)

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

from sympy import sqrt, sin, cos, tan, pi, pprint

print('6.')
for op in [sin, cos, tan]:
    pprint(op(pi / 12))

print('7.')
for op in [sin, cos, tan]:
    pprint(op(pi / 8))

入出力結果(Terminal, IPython)

$ ./sample6.py
6.
  √2   √6
- ── + ──
  4    4 
√2   √6
── + ──
4    4 
-√3 + 2
7.
    __________
   ╱   √2   1 
  ╱  - ── + ─ 
╲╱     4    2 
    ________
   ╱ √2   1 
  ╱  ── + ─ 
╲╱   4    2 
   ⎛  √2    ⎞
√2⋅⎜- ── + 1⎟
   ⎝  2     ⎠
$