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









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
















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









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














5












5








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









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. 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




trigonometry






share|cite|improve this question









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









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




share|cite|improve this question








edited 1 hour ago









N. F. Taussig

45.5k103358




45.5k103358






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




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.





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.






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








  • 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














  • 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








1




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






$begingroup$
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




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






$begingroup$
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




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










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 $-frac{pi}{2}leq x leq frac{pi}{2}$, or $-90^{circ} leq x leq 90^{circ}$ in degree mode, as seen in the following plot:



enter image description here



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.



enter image description here



Lastly, for $tan(x)$ we can take a full period around the origin, so $-frac{pi}{2} leq x leq frac{pi}{2}$, or $-90^{circ} leq x leq 90^{circ}$ in degree mode.



enter image description here






share|cite|improve this answer









$endgroup$









  • 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
    1 hour ago












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


















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















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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 $-frac{pi}{2}leq x leq frac{pi}{2}$, or $-90^{circ} leq x leq 90^{circ}$ in degree mode, as seen in the following plot:



enter image description here



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.



enter image description here



Lastly, for $tan(x)$ we can take a full period around the origin, so $-frac{pi}{2} leq x leq frac{pi}{2}$, or $-90^{circ} leq x leq 90^{circ}$ in degree mode.



enter image description here






share|cite|improve this answer









$endgroup$









  • 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
    1 hour ago
















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 $-frac{pi}{2}leq x leq frac{pi}{2}$, or $-90^{circ} leq x leq 90^{circ}$ in degree mode, as seen in the following plot:



enter image description here



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.



enter image description here



Lastly, for $tan(x)$ we can take a full period around the origin, so $-frac{pi}{2} leq x leq frac{pi}{2}$, or $-90^{circ} leq x leq 90^{circ}$ in degree mode.



enter image description here






share|cite|improve this answer









$endgroup$









  • 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
    1 hour ago














3












3








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 $-frac{pi}{2}leq x leq frac{pi}{2}$, or $-90^{circ} leq x leq 90^{circ}$ in degree mode, as seen in the following plot:



enter image description here



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.



enter image description here



Lastly, for $tan(x)$ we can take a full period around the origin, so $-frac{pi}{2} leq x leq frac{pi}{2}$, or $-90^{circ} leq x leq 90^{circ}$ in degree mode.



enter image description here






share|cite|improve this answer









$endgroup$



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 $-frac{pi}{2}leq x leq frac{pi}{2}$, or $-90^{circ} leq x leq 90^{circ}$ in degree mode, as seen in the following plot:



enter image description here



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.



enter image description here



Lastly, for $tan(x)$ we can take a full period around the origin, so $-frac{pi}{2} leq x leq frac{pi}{2}$, or $-90^{circ} leq x leq 90^{circ}$ in degree mode.



enter image description here







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 1 hour ago









DMcMorDMcMor

2,96821328




2,96821328








  • 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
    1 hour ago














  • 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
    1 hour ago








2




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
1 hour ago




$begingroup$
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
1 hour ago










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










draft saved

draft discarded


















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













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












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
















Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














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















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

What other Star Trek series did the main TNG cast show up in?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75 ...
Read more

Berlina muro

Image
La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l...
Read more

Berlina aerponto

Image
Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa...
Read more