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1 /* -*- Mode: C++; tab-width: 2; indent-tabs-mode: nil; c-basic-offset: 2 -*- */ |
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2 /* This Source Code Form is subject to the terms of the Mozilla Public |
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3 * License, v. 2.0. If a copy of the MPL was not distributed with this |
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4 * file, You can obtain one at http://mozilla.org/MPL/2.0/. */ |
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5 |
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6 |
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7 /** |
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8 File Name: 15.8.2.18.js |
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9 ECMA Section: 15.8.2.18 tan( x ) |
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10 Description: return an approximation to the tan of the |
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11 argument. argument is expressed in radians |
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12 special cases: |
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13 - if x is NaN result is NaN |
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14 - if x is 0 result is 0 |
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15 - if x is -0 result is -0 |
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16 - if x is Infinity or -Infinity result is NaN |
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17 Author: christine@netscape.com |
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18 Date: 7 july 1997 |
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19 */ |
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20 |
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21 var SECTION = "15.8.2.18"; |
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22 var VERSION = "ECMA_1"; |
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23 startTest(); |
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24 var TITLE = "Math.tan(x)"; |
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25 var EXCLUDE = "true"; |
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26 |
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27 writeHeaderToLog( SECTION + " "+ TITLE); |
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28 |
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29 new TestCase( SECTION, |
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30 "Math.tan.length", |
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31 1, |
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32 Math.tan.length ); |
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33 |
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34 new TestCase( SECTION, |
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35 "Math.tan()", |
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36 Number.NaN, |
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37 Math.tan() ); |
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38 |
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39 new TestCase( SECTION, |
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40 "Math.tan(void 0)", |
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41 Number.NaN, |
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42 Math.tan(void 0)); |
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43 |
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44 new TestCase( SECTION, |
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45 "Math.tan(null)", |
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46 0, |
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47 Math.tan(null) ); |
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48 |
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49 new TestCase( SECTION, |
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50 "Math.tan(false)", |
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51 0, |
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52 Math.tan(false) ); |
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53 |
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54 new TestCase( SECTION, |
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55 "Math.tan(NaN)", |
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56 Number.NaN, |
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57 Math.tan(Number.NaN) ); |
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58 |
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59 new TestCase( SECTION, |
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60 "Math.tan(0)", |
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61 0, |
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62 Math.tan(0)); |
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63 |
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64 new TestCase( SECTION, |
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65 "Math.tan(-0)", |
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66 -0, |
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67 Math.tan(-0)); |
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68 |
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69 new TestCase( SECTION, |
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70 "Math.tan(Infinity)", |
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71 Number.NaN, |
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72 Math.tan(Number.POSITIVE_INFINITY)); |
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73 |
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74 new TestCase( SECTION, |
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75 "Math.tan(-Infinity)", |
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76 Number.NaN, |
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77 Math.tan(Number.NEGATIVE_INFINITY)); |
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78 |
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79 new TestCase( SECTION, |
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80 "Math.tan(Math.PI/4)", |
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81 1, |
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82 Math.tan(Math.PI/4)); |
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83 |
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84 new TestCase( SECTION, |
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85 "Math.tan(3*Math.PI/4)", |
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86 -1, |
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87 Math.tan(3*Math.PI/4)); |
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88 |
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89 new TestCase( SECTION, |
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90 "Math.tan(Math.PI)", |
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91 -0, |
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92 Math.tan(Math.PI)); |
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93 |
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94 new TestCase( SECTION, |
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95 "Math.tan(5*Math.PI/4)", |
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96 1, |
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97 Math.tan(5*Math.PI/4)); |
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98 |
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99 new TestCase( SECTION, |
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100 "Math.tan(7*Math.PI/4)", |
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101 -1, |
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102 Math.tan(7*Math.PI/4)); |
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103 |
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104 new TestCase( SECTION, |
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105 "Infinity/Math.tan(-0)", |
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106 -Infinity, |
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107 Infinity/Math.tan(-0) ); |
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108 |
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109 /* |
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110 Arctan (x) ~ PI/2 - 1/x for large x. For x = 1.6x10^16, 1/x is about the last binary digit of double precision PI/2. |
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111 That is to say, perturbing PI/2 by this much is about the smallest rounding error possible. |
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112 |
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113 This suggests that the answer Christine is getting and a real Infinity are "adjacent" results from the tangent function. I |
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114 suspect that tan (PI/2 + one ulp) is a negative result about the same size as tan (PI/2) and that this pair are the closest |
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115 results to infinity that the algorithm can deliver. |
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116 |
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117 In any case, my call is that the answer we're seeing is "right". I suggest the test pass on any result this size or larger. |
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118 = C = |
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119 */ |
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120 |
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121 new TestCase( SECTION, |
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122 "Math.tan(3*Math.PI/2) >= 5443000000000000", |
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123 true, |
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124 Math.tan(3*Math.PI/2) >= 5443000000000000 ); |
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125 |
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126 new TestCase( SECTION, |
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127 "Math.tan(Math.PI/2) >= 5443000000000000", |
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128 true, |
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129 Math.tan(Math.PI/2) >= 5443000000000000 ); |
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130 |
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131 test(); |