数学 - Python - JavaScript - 解析学 - 微分と基本的な関数 - 正弦と余弦 - 加法公式(分解、既知の値)

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
参考書籍

解析入門原書第3版 (S.ラング(著)、松坂和夫(翻訳)、片山孝次(翻訳)、岩波書店)の第2部(微分と基本的な関数)、第4章(正弦と余弦)、3(加法公式)、練習問題1、2、3.を取り組んでみる。

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

コード(Emacs)

Python 3

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

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


for x in [7 * pi / 12, pi / 12, 5 * pi / 12, 11 * pi / 12, 10 * pi / 12]:
    pprint(x)
    print('sin, cos')
    pprint(factor(sin(x)))
    pprint(factor(cos(x)))
    print()

入出力結果(Terminal, IPython)

$ ./sample1.py
7⋅π
───
 12
sin, cos
√2 + √6
───────
   4   
-√6 + √2
────────
   4    

π 
──
12
sin, cos
-(-√6 + √2) 
────────────
     4      
√2 + √6
───────
   4   

5⋅π
───
 12
sin, cos
√2 + √6
───────
   4   
-(-√6 + √2) 
────────────
     4      

11⋅π
────
 12 
sin, cos
-(-√6 + √2) 
────────────
     4      
-(√2 + √6) 
───────────
     4     

5⋅π
───
 6 
sin, cos
1/2
-√3 
────
 2  
$

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="0.001">
<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">
<br>
<label for="x0">x0 = </label>
<input id="x0" type="number" value="1">

<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="sample1.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'),
    input_x0 = document.querySelector('#x0'),
    inputs = [input_r, input_dx, input_x1, input_x2, input_y1, input_y2,
             input_x0],    
    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 f = (x) => x ** 2 * Math.sin(1 / x),
    g = (x) => x * Math.sin(1 / x);

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),
        x0 = parseFloat(input_x0.value);
    
    if (r === 0 || dx === 0 || x1 > x2 || y1 > y2) {
        return;n
    }

    let points = [],
        lines = [[x1, Math.sin(x0), x2, Math.sin(x0)],
                 [x1, Math.cos(x0), x2, Math.cos(x0)],
                 [x0, y1, x0, y2]];
    
    for (let x = x1; x <= x2; x += dx) {
        let y = Math.sin(x);

        if (Math.abs(y) < Infinity) {
            points.push([x, y]);
        }
    }
    let t = points.length;
    
    for (let x = x1; x <= x2; x += dx) {
        let y = Math.cos(x);

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

    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', '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, i) =>
              i < t ? 'green' : 'blue');
    
    svg.append('g')
        .attr('transform', `translate(0, ${height - padding})`)
        .call(xaxis);

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

    p(`sin(${x0}) = ${Math.sin(x0)}, cos(${x0}) = ${Math.cos(x0)}`);
};

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

r = dx =
x1 = x2 =
y1 = y2 =
x0 =

Kamimura's blog

2017年6月6日火曜日

数学 - Python - JavaScript - 解析学 - 微分と基本的な関数 - 正弦と余弦 - 加法公式(分解、既知の値)

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