函數繪圖 測試

[狐鬼瀟湘]

代碼: 選擇全部

var board = JXG.JSXGraph.initBoard('jxgbox', {boundingbox: [-4, 3, 4, -3]});
            board.options.label.autoPosition = true;
            board.options.point.size = 1;
var A = board.create('point' , [-1.2,-2], {color:  'orange' , size: 4 });
var B = board.create('point',  [0.25,-0.5], {color:  'orange' , size: 4 });
var hexagon = board.create('regularpolygon', [A,B, 6]);
var D = hexagon.vertices[3];
var Q = board.create('circumcenter',  [A, B, D], {name:'Q'});
var G = board.create('point', [3,-2],  {name: 'G', color: 'orange' , size: 4 });
var rtr = board.create('regularpolygon',  [B, G, 3]);
var H = rtr.vertices[2];
var R = board.create('circumcenter',  [B, G, H], {name: 'R'});
var tr = board.create('polygon',  [A, G, B], {color: 'pink'});
var P = board.create('midpoint', [A, G], {name: 'P'});
var q = board.create('line', [P, Q], {name: 'q', withLabel: true});
var r = board.create('line', [P, R], {name: 'r', withLabel: true});
var angle = board.create('angle', [R, P, Q], {radius: 0.4, color: 'red', fillOpacity: 0 , name:'ϕ' });
            board.create('text', [-3, -3, 
                function () {return 'θ_1 = ' + (arc2.Value() * 180 /Math.PI).toFixed(1) + '°';}
                ]); 

[狐鬼瀟湘]

\(\begin{eqnarray}
16A^2&=&4a^2b^2+4c^2d^2-(a^2+b^2-c^2-d^2)^2-8abcd\cos(\theta+\phi) \\ \\
　　&=&4a^2b^2+4c^2d^2+8abcd-(a^2+b^2-c^2-d^2)^2-8abcd\left [1+\cos(\theta+\phi)\right ] \\ \\
　　&=&(2ab+2cd)^2-(a^2+b^2-c^2-d^2)^2-8abcd\left [2\cos^2(\frac{\theta+\phi}{2})\right ]　←\cos2x=2cos^2x-1 \\ \\
　　&=&(2ab+2cd+a^2+b^2-c^2-d^2)(2ab+2cd-a^2-b^2+c^2+d^2)-16abcd\cos^2(\frac{\theta+\phi}{2})\\ \\
　　&=&\left [(a+b)^2-(c-d)^2 \right ]\left [(c+d)^2-(a-b)^2 \right ]-16abcd\cos^2(\frac{\theta+\phi}{2}) \\ \\
　　&=&(a+b+c-d)(a+b+d-c)(a+c+d-b)(b+c+d-a)-16abcd\cos^2(\frac{\theta+\phi}{2}) \\ \\
　　&=&16(\frac{b+c+d-a}{2})(\frac{a+c+d-b}{2})(\frac{a+b+d-c}{2})(\frac{a+b+c-d}{2})-16abcd\cos^2(\frac{\theta+\phi}{2}) \\ \\
　　&=&16(\frac{a+b+c+d}{2}-a)(\frac{a+b+c+d}{2}-b)(\frac{a+b+c+d}{2}-c)(\frac{a+b+c+d}{2}-d)-16abcd\cos^2(\frac{\theta+\phi}{2})\end{eqnarray}\)

代碼: 選擇全部

[latex]\begin{eqnarray}
16A^2&=&4a^2b^2+4c^2d^2-(a^2+b^2-c^2-d^2)^2-8abcd\cos(\theta+\phi) \\ \\
　　&=&4a^2b^2+4c^2d^2+8abcd-(a^2+b^2-c^2-d^2)^2-8abcd\left [1+\cos(\theta+\phi)\right ] \\ \\
　　&=&(2ab+2cd)^2-(a^2+b^2-c^2-d^2)^2-8abcd\left [2\cos^2(\frac{\theta+\phi}{2})\right ]　←\cos2x=2cos^2x-1 \\ \\
　　&=&(2ab+2cd+a^2+b^2-c^2-d^2)(2ab+2cd-a^2-b^2+c^2+d^2)-16abcd\cos^2(\frac{\theta+\phi}{2})\\ \\
　　&=&\left [(a+b)^2-(c-d)^2  \right ]\left [(c+d)^2-(a-b)^2  \right ]-16abcd\cos^2(\frac{\theta+\phi}{2}) \\ \\
　　&=&(a+b+c-d)(a+b+d-c)(a+c+d-b)(b+c+d-a)-16abcd\cos^2(\frac{\theta+\phi}{2}) \\ \\
　　&=&16(\frac{b+c+d-a}{2})(\frac{a+c+d-b}{2})(\frac{a+b+d-c}{2})(\frac{a+b+c-d}{2})-16abcd\cos^2(\frac{\theta+\phi}{2}) \\ \\
　　&=&16(\frac{a+b+c+d}{2}-a)(\frac{a+b+c+d}{2}-b)(\frac{a+b+c+d}{2}-c)(\frac{a+b+c+d}{2}-d)-16abcd\cos^2(\frac{\theta+\phi}{2})\end{eqnarray}[/latex]

\(\color{#00000000}{4a^2b^2+4c^2d^2+8abcd}-(a^2+b^2-c^2-d^2)^2-8abcd\left [1+\cos(\theta+\phi)\right ]\)

代碼: 選擇全部

[latex]\color{#00000000}{4a^2b^2+4c^2d^2+8abcd}-(a^2+b^2-c^2-d^2)^2-8abcd\left [1+\cos(\theta+\phi)\right ][/latex]