Mfcttrf

Weaponising the Grasp-at-a-Distance spell

Random body shuffle every night—can we still function?

Is it OK to use the testing sample to compare algorithms?

My mentor says to set image to Fine instead of RAW — how is this different from JPG?

Centre cell vertically in tabularx

Can gravitational waves pass through a black hole?

.bashrc alias for a command with fixed second parameter

Did pre-Columbian Americans know the spherical shape of the Earth?

Determine whether an integer is a palindrome

The Nth Gryphon Number

Why did Bronn offer to be Tyrion Lannister's champion in trial by combat?

Is this Kuo-toa homebrew race balanced?

Why does BitLocker not use RSA?

Why complex landing gears are used instead of simple, reliable and light weight muscle wire or shape memory alloys?

Did John Wesley plagiarize Matthew Henry...?

Besides transaction validation, are there any other uses of the Script language in Bitcoin

Inverse square law not accurate for non-point masses?

Does the universe have a fixed centre of mass?

Problem with display of presentation

How to achieve cat-like agility?

Is there a spell that can create a permanent fire?

How to ask rejected full-time candidates to apply to teach individual courses?

Marquee sign letters

How to make triangles with rounded sides and corners? (squircle with 3 sides)

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 1 hour ago

N. F. Taussig

45.5k103358

asked 2 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
  2 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
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This tutorial explains how to typeset mathematics on this site.
  $endgroup$
  – N. F. Taussig
  1 hour 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 1 hour ago

N. F. Taussig

45.5k103358

asked 2 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
  2 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
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This tutorial explains how to typeset mathematics on this site.
  $endgroup$
  – N. F. Taussig
  1 hour 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 1 hour ago

N. F. Taussig

45.5k103358

asked 2 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 1 hour ago

N. F. Taussig

45.5k103358

asked 2 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 1 hour ago

N. F. Taussig

45.5k103358

asked 2 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 1 hour ago

N. F. Taussig

45.5k103358

edited 1 hour ago

N. F. Taussig

45.5k103358

asked 2 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 2 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

2,96821328

$endgroup$

• 
  
  
  

  
  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');
);

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

2,96821328

$endgroup$

• 
  
  
  

  
  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

2,96821328

$endgroup$

• 
  
  
  

  
  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

2,96821328

$endgroup$

share|cite|improve this answer

answered 2 hours ago

DMcMorDMcMor

2,96821328

answered 2 hours ago

DMcMorDMcMor

2,96821328

answered 2 hours ago

DMcMorDMcMor

2,96821328