| /* -*- 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 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(); |