[从头学数学] 第213节 带着计算机去高考（五）

剧情提要：
[机器小伟]在[工程师阿伟]的陪同下进入了[九转金丹]之第八转的修炼。设想一个场景：
如果允许你带一台不连网的计算机去参加高考，你会放弃选择一个手拿计算器和草稿本吗
？阿伟决定和小伟来尝试一下用计算机算高考题会是怎样的感觉。

正剧开始：

星历2016年05月19日 16:36:52, 银河系厄尔斯星球中华帝国江南行省。
[工程师阿伟]正在和[机器小伟]一起做着2004年的江苏省数学高考题]。

总而言之，上一年的难度让考生不满意，这一次就果断下来了。

<span style="font-size:18px;">//题2
	if (1) {      
        var r = 20;            
        config.setSector(3,1,2.5,1);              
        config.graphPaper2D(0, 0, r);            
        config.axis2D(0, 0,180);              
                
        //坐标轴设定        
        var scaleX = 2*r, scaleY = 2*r;          
        var spaceX = 45, spaceY = 1;           
        var xS = -360, xE = 360;          
        var yS = -10, yE = 10;          
        config.axisSpacing(xS, xE, spaceX, scaleX, 'X');            
        config.axisSpacing(yS, yE, spaceY, scaleY, 'Y');            
                    
        var transform = new Transform();            
        //存放函数图像上的点        
        var a = [], b = [], c = [], d = [];          
                  
        //需要显示的函数说明    
                    //希腊字母表(存此用于Ctrl C/V  
            //ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ  
            //αβγδεζηθικλμνξοπρστυφχψω  
        var f1 = 'y = 2cos^2(x)+1', f2 = '', f3 = '', f4 = '';        
        var  y1 = y2 = 0;
		//(1-(x+2)^2)^0.5
		for (var x = xS; x <= xE; x+=10) { 
				y = taskFun(x);
				a.push([x, y]);

        }  
          
                  
        //存放临时数组        
        var tmp = [];        
                  
        //显示变换        
        if (a.length > 0) {        
            a = transform.scale(transform.translate(a, 0, 0), scaleX/spaceX, scaleY/spaceY);         
            //函数1        
            tmp = [].concat(a);            
            shape.pointDraw(tmp, 'red');            
           
			tmp = [].concat(a);            
            shape.multiLineDraw(tmp, 'pink');        
                      
            plot.setFillStyle('red');        
            plot.fillText(f1, 100, -90, 200);          
        }  
		


    } 
	

}  

function taskFun(x) {
	return 2*Math.pow(Math.cos(x/180*Math.PI), 2) + 1;
}
</span>

<span style="font-size:18px;">#题3
def tmp3():
    sum_ = 0;
    for i in range(1, 4):
        a = alg.combination(4, i)*alg.combination(3, 4-i);
        print(a);
        sum_ += a;

    print('sum = ', sum_);</span>

<span style="font-size:18px;">#题7
def tmp7():
    A = ['2x', 'x^[0.5]'];

    A = alg.strformat(A);
    A = alg.strpow_n(A, 4);
    A = alg.strcombine(A);

    print(A);</span>

>>> 
['(16)*x^[4]', '(32)*x^[3.5]', '(24)*x^[3.0]', '(8)*x^[2.5]', '(1)*x^[2.0]']

<span style="font-size:18px;">//题10
	if (1) {      
        var r = 20;            
        config.setSector(1,1,1,1);              
        config.graphPaper2D(0, 0, r);            
        config.axis2D(0, 0,180);              
                
        //坐标轴设定        
        var scaleX = 2*r, scaleY = 2*r;          
        var spaceX = 1, spaceY = 5;           
        var xS = -10, xE = 10;          
        var yS = -20, yE = 20;          
        config.axisSpacing(xS, xE, spaceX, scaleX, 'X');            
        config.axisSpacing(yS, yE, spaceY, scaleY, 'Y');            
                    
        var transform = new Transform();            
        //存放函数图像上的点        
        var a = [], b = [], c = [], d = [];          
                  
        //需要显示的函数说明    
                    //希腊字母表(存此用于Ctrl C/V  
            //ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩ  
            //αβγδεζηθικλμνξοπρστυφχψω  
        var f1 = 'y = x^3- 3*x+1', f2 = '', f3 = '', f4 = '';        
        var  y1 = y2 = 0;
		//(1-(x+2)^2)^0.5
		for (var x = xS; x <= xE; x+=0.2) { 
				y = taskFun(x);
				a.push([x, y]);					

        }  
          
                  
        //存放临时数组        
        var tmp = [];        
                  
        //显示变换        
        if (a.length > 0) {        
            a = transform.scale(transform.translate(a, 0, 0), scaleX/spaceX, scaleY/spaceY);         
            //函数1        
            tmp = [].concat(a);            
            shape.pointDraw(tmp, 'red');            
           
			tmp = [].concat(a);            
            shape.multiLineDraw(tmp, 'pink');          
                      
            plot.setFillStyle('red');        
            plot.fillText(f1, 100, -90, 200);          
        }  
		


    } </span>

<span style="font-size:18px;">//题17
	if (1) {
		var equation = new Equation();  
          
        var a = [1, -5/2, 1] 
        equation.quadratic(a);  
	}

方程 1x^[2] + -2.5x + 1 = 0 =>
Δ = b^[2] - 4ac = 2.25 ;
方程的解为：x1 = 0.5, x2 = 2 ;
方程根与系数的关系：x1 + x2 = 2.5, x1 * x2 = 1 ;

#题17
def tmp17():
    a = 2*math.atan(0.5);
    print(a);
    print(math.sin(a-math.pi/3));
	
>>> 
0.9272952180016122
-0.11961524227066311</span>

<span style="font-size:18px;">#题18
def tmp18():
    #三角形BCP
    BCP = geo.solveTriangle([1, '?', 4, '?', 90, '?']);

    print(BCP);#[1, 4.123105625617661, 4, 14.036243467926484, 90.0, 75.96375653207353]
    BP = 4.123

    #三角形ABC
    ABC = [4, 4*1.414, 4, '?', '?', '?'];

    #三角形ACP
    ACP = [1, '?', ABC[1], '?', 90, '?'];
    ACP = geo.solveTriangle(ACP);

    #三角形ABP
    ABP = [BP, ACP[1], 4, '?', 90, '?'];
    ABP = geo.solveTriangle(ABP);
    print(ABP[5]);


>>> 
[1, 4.123105625617661, 4, 14.036243467926484, 90.0, 75.96375653207353]
44.13988858744975
>>> math.atan(4/(17)**0.5)*180/math.pi
44.131749212031046</span>

送分题还是满多的。

本节到此结束，欲知后事如何，请看下回分解。