剧情提要：
[机器小伟]在[工程师阿伟]的陪同下进入了结丹中期的修炼，
这次要修炼的目标是[三角初等变换]。
正剧开始：
星历2016年04月15日 15:32:35, 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]一起研究[三角初等变换]。
<span style="font-size:18px;"> if (1) {
var mathText = new MathText();
var s = [
'和角、差角公式',
'sin(A+B) = sinAcosB+sinBcosA',
'sin(A-B) = sinAcosB-sinBcosA',
' ',
'cos(A+B) = cosAcosB-sinAsinB',
'cos(A-B) = cosAcosB+sinAsinB',
' ',
'tan(A+B) = (tanA+tanB)/(1-tanA*tanB)',
'tan(A-B) = (tanA-tanB)/(1+tanA*tanB)',
];
var x =40, y=40;
var r1 = 40;
var len = s.length;
for (var i = 0; i < len; i++) {
if (s[i] == '') {
if (x < 100) {
x += 300;
y-=r1*3;
}
else {
x = 20;
y += r1;
}
}
else {
mathText.print(s[i], x, y);
y+=r1;
}
}
}</span>
<span style="font-size:18px;"> var s = [ '倍角公式', 'sin2a = 2sinacosa', 'cos2a = cos^[2]a-sin^[2]a', 'tan2a = 2tana/(1-tan^[2]a)', ' ', 'cos2a = 2cos^[2]a-1', 'cos2a = 1-2sin^[2]a', ];</span>
<span style="font-size:18px;"> var s = [ '半角公式', 'sin(a/2) = ±((1-cosa)/2)^[0.5]', 'cos(a/2) = ±((1+cosa)/2)^[0.5]', 'tan(a/2) = ±((1-cosa)/(1+cosa))^[0.5] ', '= sina/(1+cosa) ', '= (1-cosa)/sina', ]; </span>
以前这种公式还有很多，都是前人的经验和智囊的结晶，
可是时过境迁哪，现在不管怎么算，哪比得上先求出角度，
再直接算来得准和快呢。所以[人叫板老师]也觉得这些公式
点到为止就行了。
当然，小伟还是做了一个工具，一般来说应该够用了：
<span style="font-size:18px;">>>>
sin45° = 0.707
cos45° = 0.707
tan45° = 1.0
sin30° = 0.5
cos30° = 0.866
tan30° = 0.577
sin75° = 0.966
cos75° = 0.259
tan75° = 3.732
sin15° = 0.259
cos15° = 0.966
tan15° = 0.268
sin90° = 1.0
cos90° = 0.0
tan90° = 1.633123935319537e+16
sin135° = 0.707
cos135° = -0.707
tan135° = -1.0
sin22.5° = 0.383
cos22.5° = 0.924
tan22.5° = 0.414
#三角恒等式
def tmp(angle1, angle2 = 0):
task = [
#第一个角的三角函数值
['sin', angle1],
['cos', angle1],
['tan', angle1],
#第二个角的三角函数值
['sin', angle2],
['cos', angle2],
['tan', angle2],
#和角公式
['sin', angle1+angle2],
['cos', angle1+angle2],
['tan', angle1+angle2],
#差角公式
['sin', angle1-angle2],
['cos', angle1-angle2],
['tan', angle1-angle2],
#二倍角
['sin', angle1*2],
['cos', angle1*2],
['tan', angle1*2],
#三倍角
['sin', angle1*3],
['cos', angle1*3],
['tan', angle1*3],
#半角
['sin', angle1/2],
['cos', angle1/2],
['tan', angle1/2]
];
triFun(task);
if __name__ == '__main__':
tmp(45, 30);</span>
<span style="font-size:18px;">>>>
5400
[0.0002908882045634, 0.00058178, 0.00087266, 0.00116355, 0.00145444, 0.00174533, 0.00203622, 0.0023271, 0.00261799, 0.00290888, 0.00319976, 0.00349065, 0.00378154, 0.00407242, 0.00436331, 0.00465419, 0.00494508, 0.00523596, 0.00552685, 0.00581773, 0.00610861, 0.0063995, 0.00669038, 0.00698126, 0.00727214, 0.00756302, 0.0078539, 0.00814478, 0.00843566, 0.00872654, 0.00901741, 0.00930829, 0.00959916, 0.00989004, 0.01018091, 0.01047178, 0.01076266, 0.01105353, 0.0113444, 0.01163527, 0.01192613, 0.012217, 0.01250787, 0.01279873, 0.0130896, 0.01338046, 0.01367132, 0.01396218, 0.01425304, 0.0145439, 0.01483475, 0.01512561, 0.01541646, 0.01570732, 0.01599817, 0.01628902, 0.01657987, 0.01687072, 0.01716156, 0.01745241, 0.01774325, 0.01803409, 0.01832493, 0.01861577, 0.01890661, 0.01919744, 0.01948828, 0.01977911, 0.02006994, 0.02036077, 0.02065159, 0.02094242, 0.02123324, 0.02152407, 0.02181489, 0.0221057, 0.02239652, 0.02268733, 0.02297815, 0.02326896, 0.02355976, 0.02385057, 0.02414138, 0.02443218, 0.02472298, 0.02501378, 0.02530457, 0.02559537, 0.02588616, 0.02617695, 0.02646774, 0.02675852, 0.0270493, 0.02734008, 0.02763086, 0.02792164, 0.02821241, 0.02850318, 0.02879395, 0.02908472]
#[0, 90]度角范围内正弦三角函数表
def triTable():
import math;
#1'角的三角函数值
sin1 = 0.0002908882045634;
cos1 = math.sqrt(1-sin1*sin1);
#90度内
sin = sin1;
cos = cos1;
#存放值
array = [];
array.append(sin);
while (sin < 1):
sin = sin1*cos+cos1*sin;
cos = math.sqrt(1-sin*sin);
array.append(round(sin, 8));
#每次1'角递增，array数组应有元素90*60=5400个
print(len(array));
#看看前100个元素
print(array[:100]);</span>
<span style="font-size:18px;">###
# @usage 三角函数调用封装
# @author mw
# @date 2016年04月05日 星期二 09:36:50
# @param
# @return
#
###
def triFun(task, prec = 3):
#需要计算的任务
#输入量为角度制，依次为度，分，秒
#可输出显示度、分、秒的表达式
#用于计算锐角三角函数
#对于角度变换，还没有经过测试
#以后可扩展添加其它多种数学函数
size = len(task);
result = [];
import math;
precision = max(prec, 3);
for i in range(size):
if task[i][0] == 'sin' or task[i][0] == 'cos' or task[i][0] == 'tan' or task[i][0] == 'cot':
taskNew = task[i];
part1 = angleTransfer(taskNew[1:]);
len1 = len(part1);
taskNew = [taskNew[0]];
for n in range(len1):
taskNew.append(part1[n]);
size_1 = len(task[i]);
angle = 0;
if size_1 == 2:
angle = taskNew[1];
elif size_1 == 3:
angle = taskNew[1]+taskNew[2]/60;
elif size_1 == 4:
angle = taskNew[1]+taskNew[2]/60+taskNew[3]/3600;
angle = angle/180*math.pi;
if task[i][0] == 'sin':
s = '';
if size_1 == 2:
s = task[i][0]+str(task[i][1])+'°';
elif size_1 == 3:
s = task[i][0]+str(task[i][1])+'°'+str(task[i][2])+'\'';
elif size_1 == 4:
s = task[i][0]+str(task[i][1])+'°'+str(task[i][2])+'\''+str(task[i][3])+'″';
s += ' = ';
s += str(round(math.sin(angle), precision));
s += '\n';
result.append(s);
elif task[i][0] == 'cos':
s = '';
if size_1 == 2:
s = task[i][0]+str(task[i][1])+'°';
elif size_1 == 3:
s = task[i][0]+str(task[i][1])+'°'+str(task[i][2])+'\'';
elif size_1 == 4:
s = task[i][0]+str(task[i][1])+'°'+str(task[i][2])+'\''+str(task[i][3])+'″';
s += ' = ';
s += str(round(math.cos(angle), precision));
s += '\n';
result.append(s);
elif task[i][0] == 'tan':
s = '';
if size_1 == 2:
s = task[i][0]+str(task[i][1])+'°';
elif size_1 == 3:
s = task[i][0]+str(task[i][1])+'°'+str(task[i][2])+'\'';
elif size_1 == 4:
s = task[i][0]+str(task[i][1])+'°'+str(task[i][2])+'\''+str(task[i][3])+'″';
s += ' = ';
s += str(round(math.tan(angle), precision));
s += '\n';
result.append(s);
elif task[i][0] == 'cot':
s = '';
if size_1 == 2:
s = task[i][0]+str(task[i][1])+'°';
elif size_1 == 3:
s = task[i][0]+str(task[i][1])+'°'+str(task[i][2])+'\'';
elif size_1 == 4:
s = task[i][0]+str(task[i][1])+'°'+str(task[i][2])+'\''+str(task[i][3])+'″';
s += ' = ';
s += str(round(1/math.tan(angle), precision));
s += '\n';
result.append(s);
else:
if task[i][0] == 'asin':
s = '';
s = task[i][0]+str(task[i][1])+' = ';
res = math.asin(task[i][1])/math.pi*180;
if res < 0:
s += '-';
absres = abs(res);
s += str(math.floor(absres))+'°'+str(math.floor(absres*60)%60)+'\''+str(round(absres*3600)%60)+'″';
s += '\n';
result.append(s);
elif task[i][0] == 'acos':
s = '';
s = task[i][0]+str(task[i][1])+' = ';
res = math.acos(task[i][1])/math.pi*180;
if res < 0:
s += '-';
absres = abs(res);
s += str(math.floor(absres))+'°'+str(math.floor(absres*60)%60)+'\''+str(round(absres*3600)%60)+'″';
s += '\n';
result.append(s);
elif task[i][0] == 'atan':
s = '';
s = task[i][0]+str(task[i][1])+' = ';
res = math.atan(task[i][1])/math.pi*180;
if res < 0:
s += '-';
absres = abs(res);
s += str(math.floor(absres))+'°'+str(math.floor(absres*60)%60)+'\''+str(round(absres*3600)%60)+'″';
s += '\n';
result.append(s);
elif task[i][0] == 'acot':
s = '';
s = task[i][0]+str(task[i][1])+' = ';
res = 90-math.atan(task[i][1])/math.pi*180;
if res < 0:
s += '-';
absres = abs(res);
s += str(math.floor(absres))+'°'+str(math.floor(absres*60)%60)+'\''+str(round(absres*3600)%60)+'″';
s += '\n';
result.append(s);
else:
pass;
size = len(result);
for i in range(size):
print(result[i]);
return result;
#把任意角转化成[0, 360]度范围内的角度。
#传入[度, 分，秒]
#传出[度, 分，秒]
def angleTransfer(a):
#传入的角度可以只有度，也可也有分，也可以有秒
#这样会造成数组的大小变动
accuracy = len(a);
if (a[0] < 0):
if accuracy == 2:
if (a[1] != 0):
a[0]-=1;
a[1] = 60-a[1];
elif accuracy == 3:
if (a[1] != 0 or a[2] != 0):
a[0] -=1;
if (a[2] != 0):
a[1] = 59-a[1];
a[2] = 60-a[2];
else:
a[1] = 60-a[1];
while a[0] < 0:
a[0]+=360;
elif (a[0] >= 360):
while a[0] >= 360:
a[0]-=360;
return a;</span>
本节到此结束，欲知后事如何，请看下回分解。
时间： 2024-04-19 11:17:13