text: /* -*- Mode: C++; tab-width: 2; indent-tabs-mode: nil; c-basic-offset: 2 -*- */
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   *
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  gTestfile = '15.8.2.18.js';
  
  /**
     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);
  
  new TestCase( SECTION,
  	      "Math.tan.length",
  	      1,
  	      Math.tan.length );
  
  new TestCase( SECTION,
  	      "Math.tan()",
  	      Number.NaN,
  	      Math.tan() );
  
  new TestCase( SECTION,
  	      "Math.tan(void 0)",
  	      Number.NaN,
  	      Math.tan(void 0));
  
  new TestCase( SECTION,
  	      "Math.tan(null)",
  	      0,
  	      Math.tan(null) );
  
  new TestCase( SECTION,
  	      "Math.tan(false)",
  	      0,
  	      Math.tan(false) );
  
  new TestCase( SECTION,
  	      "Math.tan(NaN)",
  	      Number.NaN,
  	      Math.tan(Number.NaN) );
  
  new TestCase( SECTION,
  	      "Math.tan(0)",
  	      0,
  	      Math.tan(0));
  
  new TestCase( SECTION,
  	      "Math.tan(-0)",
  	      -0,
  	      Math.tan(-0));
  
  new TestCase( SECTION,
  	      "Math.tan(Infinity)",
  	      Number.NaN,
  	      Math.tan(Number.POSITIVE_INFINITY));
  
  new TestCase( SECTION,
  	      "Math.tan(-Infinity)",
  	      Number.NaN,
  	      Math.tan(Number.NEGATIVE_INFINITY));
  
  new TestCase( SECTION,
  	      "Math.tan(Math.PI/4)",
  	      1,
  	      Math.tan(Math.PI/4));
  
  new TestCase( SECTION,
  	      "Math.tan(3*Math.PI/4)",
  	      -1,
  	      Math.tan(3*Math.PI/4));
  
  new TestCase( SECTION,
  	      "Math.tan(Math.PI)",
  	      -0,
  	      Math.tan(Math.PI));
  
  new TestCase( SECTION,
  	      "Math.tan(5*Math.PI/4)",
  	      1,
  	      Math.tan(5*Math.PI/4));
  
  new TestCase( SECTION,
  	      "Math.tan(7*Math.PI/4)",
  	      -1,
  	      Math.tan(7*Math.PI/4));
  
  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 =
  */
  
  new TestCase( SECTION,
  	      "Math.tan(3*Math.PI/2) >= 5443000000000000",
  	      true,
  	      Math.tan(3*Math.PI/2) >= 5443000000000000 );
  
  new TestCase( SECTION,
  	      "Math.tan(Math.PI/2) >= 5443000000000000",
  	      true,
  	      Math.tan(Math.PI/2) >= 5443000000000000 );
  
  test();
  