Ttykuu

What is the proper term for etching or digging of wall to hide conduit of cables

Did any compiler fully use 80-bit floating point?

Statistical analysis applied to methods coming out of Machine Learning

Centre cell vertically in tabularx

One-one communication

Keep at all times, the minus sign above aligned with minus sign below

.bashrc alias for a command with fixed second parameter

What are some likely causes to domain member PC losing contact to domain controller?

Does the universe have a fixed centre of mass?

Do i imagine the linear (straight line) homotopy in a correct way?

Is a copyright notice with a non-existent name be invalid?

How do I say "this must not happen"?

Is the Mordenkainen's Sword spell underpowered?

An isoperimetric-type inequality inside a cube

How do you cope with tons of web fonts when copying and pasting from web pages?

French equivalents of おしゃれは足元から (Every good outfit starts with the shoes)

Short story about astronauts fertilizing soil with their own bodies

Fit odd number of triplets in a measure?

When does a function NOT have an antiderivative?

Why not use the yoke to control yaw, as well as pitch and roll?

Plotting a Maclaurin series

3D Masyu - A Die

Improvising over quartal voicings

By what mechanism was the 2017 UK General Election called?

calculator's angle answer for trig ratios that can work in more than 1 quadrant on the unit circle

5

$begingroup$

1. Why does the calculator do a cc (counterclockwise) rotation for positive trig ratios instead of clockwise,




2. and a clockwise rotation for negative sine & tan instead of cc




3. and a counterclockwise rotation for negative cos ratios instead of a clockwise

ie. in degree mode

$cos^-1(-5/12)=114.62$

$sin^-1(-5/12)=-24.62$

$tan^-1(-5/12)=-22.61$

Is it maybe picking the value that involves the least amount of computing power? or is it a matter of convention? or am I overlooking something?

trigonometry

share|cite|improve this question

edited 2 hours ago

N. F. Taussig

45.5k103358

asked 3 hours ago

Allan HenriquesAllan Henriques

283

New contributor

Allan Henriques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Conventionally, counter clockwise rotations are described by positive angles. But it looks like your question is more about the ranges of the inverse trigonometric functions.
  $endgroup$
  – John Doe
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Try using Mathjax: Surround your formulas with $ signs, use before a trig function, and between the start and end of a superscript. E.g. $cos^-1(-5/12)=114.62$
  $endgroup$
  – man on laptop
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This tutorial explains how to typeset mathematics on this site.
  $endgroup$
  – N. F. Taussig
  2 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

1. Why does the calculator do a cc (counterclockwise) rotation for positive trig ratios instead of clockwise,




2. and a clockwise rotation for negative sine & tan instead of cc




3. and a counterclockwise rotation for negative cos ratios instead of a clockwise

ie. in degree mode

$cos^-1(-5/12)=114.62$

$sin^-1(-5/12)=-24.62$

$tan^-1(-5/12)=-22.61$

Is it maybe picking the value that involves the least amount of computing power? or is it a matter of convention? or am I overlooking something?

trigonometry

share|cite|improve this question

edited 2 hours ago

N. F. Taussig

45.5k103358

asked 3 hours ago

Allan HenriquesAllan Henriques

283

New contributor

Allan Henriques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Conventionally, counter clockwise rotations are described by positive angles. But it looks like your question is more about the ranges of the inverse trigonometric functions.
  $endgroup$
  – John Doe
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Try using Mathjax: Surround your formulas with $ signs, use before a trig function, and between the start and end of a superscript. E.g. $cos^-1(-5/12)=114.62$
  $endgroup$
  – man on laptop
  3 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This tutorial explains how to typeset mathematics on this site.
  $endgroup$
  – N. F. Taussig
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

1. Why does the calculator do a cc (counterclockwise) rotation for positive trig ratios instead of clockwise,




2. and a clockwise rotation for negative sine & tan instead of cc




3. and a counterclockwise rotation for negative cos ratios instead of a clockwise

ie. in degree mode

$cos^-1(-5/12)=114.62$

$sin^-1(-5/12)=-24.62$

$tan^-1(-5/12)=-22.61$

Is it maybe picking the value that involves the least amount of computing power? or is it a matter of convention? or am I overlooking something?

trigonometry

share|cite|improve this question

edited 2 hours ago

N. F. Taussig

45.5k103358

asked 3 hours ago

Allan HenriquesAllan Henriques

283

New contributor

Allan Henriques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 2 hours ago

N. F. Taussig

45.5k103358

asked 3 hours ago

Allan HenriquesAllan Henriques

283

New contributor

Allan Henriques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 hours ago

N. F. Taussig

45.5k103358

asked 3 hours ago

Allan HenriquesAllan Henriques

283

New contributor

Allan Henriques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 2 hours ago

N. F. Taussig

45.5k103358

edited 2 hours ago

N. F. Taussig

45.5k103358

asked 3 hours ago

Allan HenriquesAllan Henriques

283

New contributor

Allan Henriques is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

Allan HenriquesAllan Henriques

283

1 Answer
1

active

oldest

votes

3

$begingroup$

This actually has to do with the way inverse trig functions are defined. For a function to be invertible there must be one input for every output. Graphically, this is equivalent to passing the horizontal line test. Now, trig functions are periodic and as such are very much not invertible. The way we get around this is to restrict the domain of each function to a region that passes the horizontal line test.

For $sin(x)$ the region that we take is $-fracpi2leq x leq fracpi2$, or $-90^circ leq x leq 90^circ$ in degree mode, as seen in the following plot:

For $cos(x)$ the region we take is $0leq x leq pi$, or $0^circ leq x leq 180^circ$ in degree mode. Note that we could also have taken $-pi leq x leq 0$, but for convenience we take $x$ to be a positive angle.

Lastly, for $tan(x)$ we can take a full period around the origin, so $-fracpi2 leq x leq fracpi2$, or $-90^circ leq x leq 90^circ$ in degree mode.

share|cite|improve this answer

answered 2 hours ago

DMcMorDMcMor

2,99321328

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  That makes complete sense! When you see the graph of the functions, sure enough they give out the reasons why the calculators give out the answers they do.
  $endgroup$
  – bjcolby15
  2 hours ago
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

Allan Henriques is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');

var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);

;
$window.on('scroll', onScroll);

);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3196538%2fcalculators-angle-answer-for-trig-ratios-that-can-work-in-more-than-1-quadrant%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

This actually has to do with the way inverse trig functions are defined. For a function to be invertible there must be one input for every output. Graphically, this is equivalent to passing the horizontal line test. Now, trig functions are periodic and as such are very much not invertible. The way we get around this is to restrict the domain of each function to a region that passes the horizontal line test.

For $sin(x)$ the region that we take is $-fracpi2leq x leq fracpi2$, or $-90^circ leq x leq 90^circ$ in degree mode, as seen in the following plot:

For $cos(x)$ the region we take is $0leq x leq pi$, or $0^circ leq x leq 180^circ$ in degree mode. Note that we could also have taken $-pi leq x leq 0$, but for convenience we take $x$ to be a positive angle.

Lastly, for $tan(x)$ we can take a full period around the origin, so $-fracpi2 leq x leq fracpi2$, or $-90^circ leq x leq 90^circ$ in degree mode.

share|cite|improve this answer

answered 2 hours ago

DMcMorDMcMor

2,99321328

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  That makes complete sense! When you see the graph of the functions, sure enough they give out the reasons why the calculators give out the answers they do.
  $endgroup$
  – bjcolby15
  2 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

This actually has to do with the way inverse trig functions are defined. For a function to be invertible there must be one input for every output. Graphically, this is equivalent to passing the horizontal line test. Now, trig functions are periodic and as such are very much not invertible. The way we get around this is to restrict the domain of each function to a region that passes the horizontal line test.

For $sin(x)$ the region that we take is $-fracpi2leq x leq fracpi2$, or $-90^circ leq x leq 90^circ$ in degree mode, as seen in the following plot:

For $cos(x)$ the region we take is $0leq x leq pi$, or $0^circ leq x leq 180^circ$ in degree mode. Note that we could also have taken $-pi leq x leq 0$, but for convenience we take $x$ to be a positive angle.

Lastly, for $tan(x)$ we can take a full period around the origin, so $-fracpi2 leq x leq fracpi2$, or $-90^circ leq x leq 90^circ$ in degree mode.

share|cite|improve this answer

answered 2 hours ago

DMcMorDMcMor

2,99321328

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  That makes complete sense! When you see the graph of the functions, sure enough they give out the reasons why the calculators give out the answers they do.
  $endgroup$
  – bjcolby15
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

This actually has to do with the way inverse trig functions are defined. For a function to be invertible there must be one input for every output. Graphically, this is equivalent to passing the horizontal line test. Now, trig functions are periodic and as such are very much not invertible. The way we get around this is to restrict the domain of each function to a region that passes the horizontal line test.

For $sin(x)$ the region that we take is $-fracpi2leq x leq fracpi2$, or $-90^circ leq x leq 90^circ$ in degree mode, as seen in the following plot:

For $cos(x)$ the region we take is $0leq x leq pi$, or $0^circ leq x leq 180^circ$ in degree mode. Note that we could also have taken $-pi leq x leq 0$, but for convenience we take $x$ to be a positive angle.

Lastly, for $tan(x)$ we can take a full period around the origin, so $-fracpi2 leq x leq fracpi2$, or $-90^circ leq x leq 90^circ$ in degree mode.

share|cite|improve this answer

answered 2 hours ago

DMcMorDMcMor

2,99321328

$endgroup$

share|cite|improve this answer

answered 2 hours ago

DMcMorDMcMor

2,99321328

answered 2 hours ago

DMcMorDMcMor

2,99321328

answered 2 hours ago

DMcMorDMcMor

2,99321328