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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();
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