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+/* -*- tab-width: 2; indent-tabs-mode: nil; js-indent-level: 2 -*- */
+/* This Source Code Form is subject to the terms of the Mozilla Public
+ * License, v. 2.0. If a copy of the MPL was not distributed with this
+ * file, You can obtain one at http://mozilla.org/MPL/2.0/. */
+
+
+/**
+ 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 TITLE = "Math.tan(x)";
+var EXCLUDE = "true";
+
+writeHeaderToLog( SECTION + " "+ TITLE);
+
+new TestCase( "Math.tan.length",
+ 1,
+ Math.tan.length );
+
+new TestCase( "Math.tan()",
+ Number.NaN,
+ Math.tan() );
+
+new TestCase( "Math.tan(void 0)",
+ Number.NaN,
+ Math.tan(void 0));
+
+new TestCase( "Math.tan(null)",
+ 0,
+ Math.tan(null) );
+
+new TestCase( "Math.tan(false)",
+ 0,
+ Math.tan(false) );
+
+new TestCase( "Math.tan(NaN)",
+ Number.NaN,
+ Math.tan(Number.NaN) );
+
+new TestCase( "Math.tan(0)",
+ 0,
+ Math.tan(0));
+
+new TestCase( "Math.tan(-0)",
+ -0,
+ Math.tan(-0));
+
+new TestCase( "Math.tan(Infinity)",
+ Number.NaN,
+ Math.tan(Number.POSITIVE_INFINITY));
+
+new TestCase( "Math.tan(-Infinity)",
+ Number.NaN,
+ Math.tan(Number.NEGATIVE_INFINITY));
+
+new TestCase( "Math.tan(Math.PI/4)",
+ 1,
+ Math.tan(Math.PI/4));
+
+new TestCase( "Math.tan(3*Math.PI/4)",
+ -1,
+ Math.tan(3*Math.PI/4));
+
+new TestCase( "Math.tan(Math.PI)",
+ -0,
+ Math.tan(Math.PI));
+
+new TestCase( "Math.tan(5*Math.PI/4)",
+ 1,
+ Math.tan(5*Math.PI/4));
+
+new TestCase( "Math.tan(7*Math.PI/4)",
+ -1,
+ Math.tan(7*Math.PI/4));
+
+new TestCase( "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( "Math.tan(3*Math.PI/2) >= 5443000000000000",
+ true,
+ Math.tan(3*Math.PI/2) >= 5443000000000000 );
+
+new TestCase( "Math.tan(Math.PI/2) >= 5443000000000000",
+ true,
+ Math.tan(Math.PI/2) >= 5443000000000000 );
+
+test();