/* The contents of this file are subject to the Netscape Public * License Version 1.1 (the "License"); you may not use this file * except in compliance with the License. You may obtain a copy of * the License at http://www.mozilla.org/NPL/ * * Software distributed under the License is distributed on an "AS * IS" basis, WITHOUT WARRANTY OF ANY KIND, either express or * implied. See the License for the specific language governing * rights and limitations under the License. * * The Original Code is Mozilla Communicator client code, released March * 31, 1998. * * The Initial Developer of the Original Code is Netscape Communications * Corporation. Portions created by Netscape are * Copyright (C) 1998 Netscape Communications Corporation. All * Rights Reserved. * * Contributor(s): * */ /** File Name: 15.8.2.18.js ECMA Section: 15.8.2.18 tan( x ) Description: return an approximation to the tan of the argument. argument is expressed in radians special cases: - if x is NaN result is NaN - if x is 0 result is 0 - if x is -0 result is -0 - if x is Infinity or -Infinity result is NaN Author: christine@netscape.com Date: 7 july 1997 */ var SECTION = "15.8.2.18"; var VERSION = "ECMA_1"; startTest(); var TITLE = "Math.tan(x)"; var EXCLUDE = "true"; writeHeaderToLog( SECTION + " "+ TITLE); var testcases = getTestCases(); test(); function getTestCases() { var array = new Array(); var item = 0; array[item++] = new TestCase( SECTION, "Math.tan.length", 1, Math.tan.length ); array[item++] = new TestCase( SECTION, "Math.tan()", Number.NaN, Math.tan() ); array[item++] = new TestCase( SECTION, "Math.tan(void 0)", Number.NaN, Math.tan(void 0)); array[item++] = new TestCase( SECTION, "Math.tan(null)", 0, Math.tan(null) ); array[item++] = new TestCase( SECTION, "Math.tan(false)", 0, Math.tan(false) ); array[item++] = new TestCase( SECTION, "Math.tan(NaN)", Number.NaN, Math.tan(Number.NaN) ); array[item++] = new TestCase( SECTION, "Math.tan(0)", 0, Math.tan(0)); array[item++] = new TestCase( SECTION, "Math.tan(-0)", -0, Math.tan(-0)); array[item++] = new TestCase( SECTION, "Math.tan(Infinity)", Number.NaN, Math.tan(Number.POSITIVE_INFINITY)); array[item++] = new TestCase( SECTION, "Math.tan(-Infinity)", Number.NaN, Math.tan(Number.NEGATIVE_INFINITY)); array[item++] = new TestCase( SECTION, "Math.tan(Math.PI/4)", 1, Math.tan(Math.PI/4)); array[item++] = new TestCase( SECTION, "Math.tan(3*Math.PI/4)", -1, Math.tan(3*Math.PI/4)); array[item++] = new TestCase( SECTION, "Math.tan(Math.PI)", -0, Math.tan(Math.PI)); array[item++] = new TestCase( SECTION, "Math.tan(5*Math.PI/4)", 1, Math.tan(5*Math.PI/4)); array[item++] = new TestCase( SECTION, "Math.tan(7*Math.PI/4)", -1, Math.tan(7*Math.PI/4)); array[item++] = new TestCase( SECTION, "Infinity/Math.tan(-0)", -Infinity, Infinity/Math.tan(-0) ); /* 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. That is to say, perturbing PI/2 by this much is about the smallest rounding error possible. This suggests that the answer Christine is getting and a real Infinity are "adjacent" results from the tangent function. I 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 results to infinity that the algorithm can deliver. 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. = C = */ array[item++] = new TestCase( SECTION, "Math.tan(3*Math.PI/2) >= 5443000000000000", true, Math.tan(3*Math.PI/2) >= 5443000000000000 ); array[item++] = new TestCase( SECTION, "Math.tan(Math.PI/2) >= 5443000000000000", true, Math.tan(Math.PI/2) >= 5443000000000000 ); return ( array ); } function test() { for ( tc=0; tc < testcases.length; tc++ ) { testcases[tc].passed = writeTestCaseResult( testcases[tc].expect, testcases[tc].actual, testcases[tc].description +" = "+ testcases[tc].actual ); testcases[tc].reason += ( testcases[tc].passed ) ? "" : "wrong value "; } stopTest(); return ( testcases ); }