text: /* The contents of this file are subject to the Netscape Public License
   * Version 1.0 (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.
   * 
   */
  /**
      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 );
  }
  