michael@0: /* -*- Mode: C++; tab-width: 2; indent-tabs-mode: nil; c-basic-offset: 2 -*- */
michael@0: /* This Source Code Form is subject to the terms of the Mozilla Public
michael@0:  * License, v. 2.0. If a copy of the MPL was not distributed with this
michael@0:  * file, You can obtain one at http://mozilla.org/MPL/2.0/. */
michael@0: 
michael@0: 
michael@0: /**
michael@0:    File Name:          15.8.2.18.js
michael@0:    ECMA Section:       15.8.2.18 tan( x )
michael@0:    Description:        return an approximation to the tan of the
michael@0:    argument.  argument is expressed in radians
michael@0:    special cases:
michael@0:    - if x is NaN           result is NaN
michael@0:    - if x is 0             result is 0
michael@0:    - if x is -0            result is -0
michael@0:    - if x is Infinity or -Infinity result is NaN
michael@0:    Author:             christine@netscape.com
michael@0:    Date:               7 july 1997
michael@0: */
michael@0: 
michael@0: var SECTION = "15.8.2.18";
michael@0: var VERSION = "ECMA_1";
michael@0: startTest();
michael@0: var TITLE   = "Math.tan(x)";
michael@0: var EXCLUDE = "true";
michael@0: 
michael@0: writeHeaderToLog( SECTION + " "+ TITLE);
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan.length",
michael@0: 	      1,
michael@0: 	      Math.tan.length );
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan()",
michael@0: 	      Number.NaN,
michael@0: 	      Math.tan() );
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(void 0)",
michael@0: 	      Number.NaN,
michael@0: 	      Math.tan(void 0));
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(null)",
michael@0: 	      0,
michael@0: 	      Math.tan(null) );
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(false)",
michael@0: 	      0,
michael@0: 	      Math.tan(false) );
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(NaN)",
michael@0: 	      Number.NaN,
michael@0: 	      Math.tan(Number.NaN) );
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(0)",
michael@0: 	      0,
michael@0: 	      Math.tan(0));
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(-0)",
michael@0: 	      -0,
michael@0: 	      Math.tan(-0));
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(Infinity)",
michael@0: 	      Number.NaN,
michael@0: 	      Math.tan(Number.POSITIVE_INFINITY));
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(-Infinity)",
michael@0: 	      Number.NaN,
michael@0: 	      Math.tan(Number.NEGATIVE_INFINITY));
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(Math.PI/4)",
michael@0: 	      1,
michael@0: 	      Math.tan(Math.PI/4));
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(3*Math.PI/4)",
michael@0: 	      -1,
michael@0: 	      Math.tan(3*Math.PI/4));
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(Math.PI)",
michael@0: 	      -0,
michael@0: 	      Math.tan(Math.PI));
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(5*Math.PI/4)",
michael@0: 	      1,
michael@0: 	      Math.tan(5*Math.PI/4));
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(7*Math.PI/4)",
michael@0: 	      -1,
michael@0: 	      Math.tan(7*Math.PI/4));
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Infinity/Math.tan(-0)",
michael@0: 	      -Infinity,
michael@0: 	      Infinity/Math.tan(-0) );
michael@0: 
michael@0: /*
michael@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.
michael@0:   That is to say, perturbing PI/2 by this much is about the smallest rounding error possible.
michael@0: 
michael@0:   This suggests that the answer Christine is getting and a real Infinity are "adjacent" results from the tangent function.  I
michael@0:   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
michael@0:   results to infinity that the algorithm can deliver.
michael@0: 
michael@0:   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.
michael@0:   = C =
michael@0: */
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(3*Math.PI/2) >= 5443000000000000",
michael@0: 	      true,
michael@0: 	      Math.tan(3*Math.PI/2) >= 5443000000000000 );
michael@0: 
michael@0: new TestCase( SECTION,
michael@0: 	      "Math.tan(Math.PI/2) >= 5443000000000000",
michael@0: 	      true,
michael@0: 	      Math.tan(Math.PI/2) >= 5443000000000000 );
michael@0: 
michael@0: test();