数学 - Python - JavaScript - 三角関数(Trigonometric Function) - 三角関数の定義 - 三角関数の性質(加法定理)

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, Microsoft Edge, Google Chrome...)用JavaScript Library: MathJax
参考書籍

オイラーの贈物―人類の至宝eiπ=-1を学ぶ (吉田武(著)、東海大学出版会)の第II部(関数の定義(Definitions of Functions))、第7章(三角関数(Trigonometric Function))、7.5(三角関数の定義)、7.5.1(三角関数の性質)、問題2.を取り組んでみる。

- $\begin{array}{l} \sin (π + \frac{π}{12}) \\ = \sin π \cos \frac{π}{12} + \cos π \sin \frac{π}{12} \\ = 0 \cos \frac{π}{12} - \frac{\sqrt{6} - \sqrt{2}}{4} \\ = \frac{\sqrt{2} - \sqrt{6}}{4} \\ \cos \frac{13 π}{12} = \cos π \cos \frac{π}{12} - \sin π \sin \frac{π}{12} \\ = - \frac{\sqrt{6} + \sqrt{2}}{4} \\ \tan \frac{13 π}{12} = \frac{\tan π + \tan \frac{π}{12}}{1 - \tan π \tan \frac{π}{12}} \\ = 2 - \sqrt{3} \end{array}$
- $\begin{array}{l} \sin \frac{14 π}{12} \\ = \sin π \cos \frac{π}{6} + \cos π \sin \frac{π}{6} \\ = 0 \cos \frac{π}{6} - \frac{\sqrt{3}}{2} \\ = - \frac{\sqrt{3}}{2} \\ \cos \frac{14 π}{12} = \cos π \cos \frac{π}{6} - \sin π \sin \frac{π}{6} \\ = - \frac{1}{2} \\ \tan \frac{14 π}{12} = \frac{\tan π + \tan \frac{π}{6}}{1 - \tan π \tan \frac{π}{6}} \\ = \sqrt{3} \end{array}$
- $\begin{array}{l} \sin \frac{15 π}{12} \\ = \sin π \cos \frac{π}{4} + \cos π \sin \frac{π}{4} \\ = - \frac{1}{\sqrt{2}} \\ \cos \frac{15 π}{12} = \cos π \cos \frac{π}{4} - \sin π \sin \frac{π}{4} \\ = - \frac{1}{\sqrt{2}} \\ \tan \frac{15 π}{12} = \frac{\tan π + \tan \frac{π}{4}}{1 - \tan π \tan \frac{π}{4}} \\ = 1 \end{array}$
- $\begin{array}{l} \sin \frac{16 π}{12} \\ = \sin π \cos \frac{π}{3} + \cos π \sin \frac{π}{3} \\ = - \frac{\sqrt{3}}{2} \\ \cos \frac{16 π}{12} = \cos π \cos \frac{π}{3} - \sin π \sin \frac{π}{3} \\ = - \frac{1}{2} \\ \tan \frac{16 π}{12} = \frac{\tan π + \tan \frac{π}{3}}{1 - \tan π \tan \frac{π}{3}} \\ = \sqrt{3} \end{array}$
- $\begin{array}{l} \sin \frac{17 π}{12} \\ = \sin π \cos \frac{5 π}{12} + \cos π \sin \frac{5 π}{12} \\ = - \frac{\sqrt{6} + \sqrt{2}}{4} \\ \cos \frac{17 π}{12} = \cos π \cos \frac{5 π}{12} - \sin π \sin \frac{5 π}{12} \\ = \frac{\sqrt{2} - \sqrt{6}}{4} \\ \tan \frac{17 π}{12} = \frac{\tan π + \tan \frac{5 π}{12}}{1 - \tan π \tan \frac{5 π}{12}} \\ = 2 + \sqrt{3} \end{array}$
- $\begin{array}{l} \sin \frac{18 π}{12} \\ = \sin π \cos \frac{π}{2} + \cos π \sin \frac{π}{2} \\ = - 1 \\ \cos \frac{18 π}{12} = \cos π \cos \frac{π}{2} - \sin π \sin \frac{π}{2} \\ = 0 \\ \tan \frac{18 π}{12} = \frac{\tan π + \tan \frac{π}{2}}{1 - \tan π \tan \frac{π}{2}} \end{array}$
- $\begin{array}{l} \sin (\frac{3}{2} π + θ) \\ = \sin \frac{3 π}{2} \cos θ + \cos \frac{3 π}{2} \sin θ \\ = - \cos θ \\ \cos (\frac{3}{2} π + θ) = \cos \frac{3 π}{2} \cos θ - \sin \frac{3 π}{2} \sin θ \\ = \sin θ \\ \tan (n π + θ) = \frac{\tan n π + \tan θ}{1 - \tan n π \tan θ} \\ = \tan θ \\ - \cos \frac{π}{12} = - \frac{\sqrt{6} + \sqrt{2}}{4} \\ - \sin \frac{π}{12} = \frac{\sqrt{6} - \sqrt{2}}{4} \\ \tan \frac{7 π}{12} = - (2 + \sqrt{3}) \end{array}$
- $\begin{array}{l} \sin (π + θ) \\ = \sin π \cos θ + \cos π \sin θ \\ = - \sin θ \\ \cos (π + θ) = \cos π \cos θ - \sin π \sin θ \\ = - \cos θ \\ \tan (π + θ) = \frac{\tan π + \tan θ}{1 - \tan π \tan θ} \\ = \tan θ \\ - \sin \frac{8 π}{12} = - \frac{\sqrt{3}}{2} \\ - \cos \frac{8 π}{12} = \frac{1}{2} \\ \tan \frac{8 π}{12} = - \sqrt{3} \end{array}$
- $\begin{array}{l} \frac{9 π}{12} = \frac{3}{4} π \\ - \sin \frac{9 π}{12} = - \frac{1}{\sqrt{2}} \\ - \cos \frac{9 π}{12} = \frac{1}{\sqrt{2}} \\ \tan \frac{8 π}{12} = - 1 \end{array}$
- $\begin{array}{l} \frac{10 π}{12} = \frac{5}{6} π \\ - \sin \frac{10 π}{12} = - \frac{1}{2} \\ - \cos \frac{10 π}{12} = \frac{\sqrt{3}}{2} \\ \tan \frac{10 π}{12} = - \frac{1}{\sqrt{3}} \end{array}$
- $\begin{array}{l} - \sin \frac{11 π}{12} = \frac{\sqrt{2} - \sqrt{6}}{4} \\ - \cos \frac{11 π}{12} = \frac{\sqrt{6} + \sqrt{2}}{4} \\ \tan \frac{11 π}{12} = 2 - \sqrt{3} \end{array}$
- $\begin{array}{l} - \sin π = 0 \\ - \cos π = 1 \\ \tan π = 0 \end{array}$

コード(Emacs)

Python 3

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

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


for n in range(13, 25):
    print(n)
    pprint(list(map(lambda f: f(n * pi / 12).factor(), [sin, cos, tan])))
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample2.py
13
⎡-√6 + √2  -(√2 + √6)          ⎤
⎢────────, ───────────, -√3 + 2⎥
⎣   4           4              ⎦

14
⎡      -√3   √3⎤
⎢-1/2, ────, ──⎥
⎣       2    3 ⎦

15
⎡-√2   -√2    ⎤
⎢────, ────, 1⎥
⎣ 2     2     ⎦

16
⎡-√3           ⎤
⎢────, -1/2, √3⎥
⎣ 2            ⎦

17
⎡-(√2 + √6)   -√6 + √2        ⎤
⎢───────────, ────────, √3 + 2⎥
⎣     4          4            ⎦

18
[-1, 0, zoo]

19
⎡-(√2 + √6)   -(-√6 + √2)          ⎤
⎢───────────, ────────────, -2 - √3⎥
⎣     4            4               ⎦

20
⎡-√3           ⎤
⎢────, 1/2, -√3⎥
⎣ 2            ⎦

21
⎡-√2   √2    ⎤
⎢────, ──, -1⎥
⎣ 2    2     ⎦

22
⎡      √3  -√3 ⎤
⎢-1/2, ──, ────⎥
⎣      2    3  ⎦

23
⎡-√6 + √2  √2 + √6         ⎤
⎢────────, ───────, -2 + √3⎥
⎣   4         4            ⎦

24
[0, 1, 0]

$

HTML5

<div id="graph0"></div>
<pre id="output0"></pre>
<label for="r0">r = </label>
<input id="r0" type="number" min="0" value="0.5">
<label for="dx">dx = </label>
<input id="dx" type="number" min="0" step="0.0001" value="0.01">
<br>
<label for="x1">x1 = </label>
<input id="x1" type="number" value="-5">
<label for="x2">x2 = </label>
<input id="x2" type="number" value="5">
<br>
<label for="y1">y1 = </label>
<input id="y1" type="number" value="-5">
<label for="y2">y2 = </label>
<input id="y2" type="number" value="5">

<button id="draw0">draw</button>
<button id="clear0">clear</button>

<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/d3/4.2.6/d3.min.js" integrity="sha256-5idA201uSwHAROtCops7codXJ0vja+6wbBrZdQ6ETQc=" crossorigin="anonymous"></script>

<script src="sample2.js"></script>

JavaScript

let div0 = document.querySelector('#graph0'),
    pre0 = document.querySelector('#output0'),
    width = 600,
    height = 600,
    padding = 50,
    btn0 = document.querySelector('#draw0'),
    btn1 = document.querySelector('#clear0'),
    input_r = document.querySelector('#r0'),
    input_dx = document.querySelector('#dx'),
    input_x1 = document.querySelector('#x1'),
    input_x2 = document.querySelector('#x2'),
    input_y1 = document.querySelector('#y1'),
    input_y2 = document.querySelector('#y2'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2],
    p = (x) => pre0.textContent += x + '\n',
    range = (start, end, step=1) => {
        let res = [];
        for (let i = start; i < end; i += step) {
            res.push(i);
        }
        return res;
    };

let draw = () => {
    pre0.textContent = '';

    let r = parseFloat(input_r.value),
        dx = parseFloat(input_dx.value),
        x1 = parseFloat(input_x1.value),
        x2 = parseFloat(input_x2.value),
        y1 = parseFloat(input_y1.value),
        y2 = parseFloat(input_y2.value);

    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;
    }    

    let points = [],
        lines = [],
        fns = [[Math.sin, 'red'],
               [Math.cos, 'green'],
               [Math.tan, 'blue']],
        fns1 = [],
        fns2 = [];

    fns
        .forEach((o) => {
            let [f, color] = o;
            for (let x = x1; x <= x2; x += dx) {
                let y = f(x);

                if (Math.abs(y) < Infinity) {
                    points.push([x, y, color]);
                }
            }
        });                 

    fns2
        .forEach((o) => {
            let [f, color] = o;

            for (let x = x1; x <= x2; x += dx0) {
                let g = f(x);
                lines.push([x1, g(x1), x2, g(x2), color]);
            }
        });
    
    let xscale = d3.scaleLinear()
        .domain([x1, x2])
        .range([padding, width - padding]);
    let yscale = d3.scaleLinear()
        .domain([y1, y2])
        .range([height - padding, padding]);

    let xaxis = d3.axisBottom().scale(xscale);
    let yaxis = d3.axisLeft().scale(yscale);
    div0.innerHTML = '';
    let svg = d3.select('#graph0')
        .append('svg')
        .attr('width', width)
        .attr('height', height);

    svg.selectAll('line')
        .data([[x1, 0, x2, 0], [0, y1, 0, y2]].concat(lines))
        .enter()
        .append('line')
        .attr('x1', (d) => xscale(d[0]))
        .attr('y1', (d) => yscale(d[1]))
        .attr('x2', (d) => xscale(d[2]))
        .attr('y2', (d) => yscale(d[3]))
        .attr('stroke', (d) => d[4] || 'black');

    svg.selectAll('circle')
        .data(points)
        .enter()
        .append('circle')
        .attr('cx', (d) => xscale(d[0]))
        .attr('cy', (d) => yscale(d[1]))
        .attr('r', r)
        .attr('fill', (d) => d[2] || 'green');

    svg.append('g')
        .attr('transform', `translate(0, ${height - padding})`)
        .call(xaxis);

    svg.append('g')
        .attr('transform', `translate(${padding}, 0)`)
        .call(yaxis);

    [fns, fns1, fns2].forEach((fs) => p(fs.join('\n')));
};

inputs.forEach((input) => input.onchange = draw);
btn0.onclick = draw;
btn1.onclick = () => pre0.textContent = '';
draw();

r = dx =
x1 = x2 =
y1 = y2 =

Kamimura's blog

2017年9月24日日曜日

数学 - Python - JavaScript - 三角関数(Trigonometric Function) - 三角関数の定義 - 三角関数の性質(加法定理)

0 コメント:

コメントを投稿