15.8.2.18.js (3035B)
1 /* -*- tab-width: 2; indent-tabs-mode: nil; js-indent-level: 2 -*- */ 2 /* This Source Code Form is subject to the terms of the Mozilla Public 3 * License, v. 2.0. If a copy of the MPL was not distributed with this 4 * file, You can obtain one at http://mozilla.org/MPL/2.0/. */ 5 6 7 /** 8 File Name: 15.8.2.18.js 9 ECMA Section: 15.8.2.18 tan( x ) 10 Description: return an approximation to the tan of the 11 argument. argument is expressed in radians 12 special cases: 13 - if x is NaN result is NaN 14 - if x is 0 result is 0 15 - if x is -0 result is -0 16 - if x is Infinity or -Infinity result is NaN 17 Author: christine@netscape.com 18 Date: 7 july 1997 19 */ 20 21 var SECTION = "15.8.2.18"; 22 var TITLE = "Math.tan(x)"; 23 var EXCLUDE = "true"; 24 25 writeHeaderToLog( SECTION + " "+ TITLE); 26 27 new TestCase( "Math.tan.length", 28 1, 29 Math.tan.length ); 30 31 new TestCase( "Math.tan()", 32 Number.NaN, 33 Math.tan() ); 34 35 new TestCase( "Math.tan(void 0)", 36 Number.NaN, 37 Math.tan(void 0)); 38 39 new TestCase( "Math.tan(null)", 40 0, 41 Math.tan(null) ); 42 43 new TestCase( "Math.tan(false)", 44 0, 45 Math.tan(false) ); 46 47 new TestCase( "Math.tan(NaN)", 48 Number.NaN, 49 Math.tan(Number.NaN) ); 50 51 new TestCase( "Math.tan(0)", 52 0, 53 Math.tan(0)); 54 55 new TestCase( "Math.tan(-0)", 56 -0, 57 Math.tan(-0)); 58 59 new TestCase( "Math.tan(Infinity)", 60 Number.NaN, 61 Math.tan(Number.POSITIVE_INFINITY)); 62 63 new TestCase( "Math.tan(-Infinity)", 64 Number.NaN, 65 Math.tan(Number.NEGATIVE_INFINITY)); 66 67 new TestCase( "Math.tan(Math.PI/4)", 68 1, 69 Math.tan(Math.PI/4)); 70 71 new TestCase( "Math.tan(3*Math.PI/4)", 72 -1, 73 Math.tan(3*Math.PI/4)); 74 75 new TestCase( "Math.tan(Math.PI)", 76 -0, 77 Math.tan(Math.PI)); 78 79 new TestCase( "Math.tan(5*Math.PI/4)", 80 1, 81 Math.tan(5*Math.PI/4)); 82 83 new TestCase( "Math.tan(7*Math.PI/4)", 84 -1, 85 Math.tan(7*Math.PI/4)); 86 87 new TestCase( "Infinity/Math.tan(-0)", 88 -Infinity, 89 Infinity/Math.tan(-0) ); 90 91 /* 92 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. 93 That is to say, perturbing PI/2 by this much is about the smallest rounding error possible. 94 95 This suggests that the answer Christine is getting and a real Infinity are "adjacent" results from the tangent function. I 96 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 97 results to infinity that the algorithm can deliver. 98 99 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. 100 = C = 101 */ 102 103 new TestCase( "Math.tan(3*Math.PI/2) >= 5443000000000000", 104 true, 105 Math.tan(3*Math.PI/2) >= 5443000000000000 ); 106 107 new TestCase( "Math.tan(Math.PI/2) >= 5443000000000000", 108 true, 109 Math.tan(Math.PI/2) >= 5443000000000000 ); 110 111 test();