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
    2 hours 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
    2 hours 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
    2 hours 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 2 hours 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
    2 hours 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
    2 hours 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
2 hours 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
2 hours 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

Statuo de Libereco

Image
Koordinatoj: 40°41′21″N 74°02′40″W   /   40.68917°N , 74.04444°U  / 40.68917; -74.04444  ( Statuo de Libereco ) Statuo de Libereco angle: Statue of Liberty statuo Statuo de Libereco de antaŭe Oficiala nomo: Liberty Enlightening the World Lando  Usono Ŝtato Novjorkio Regiono Novjorko Insulo Insulo de Libereco Situo Statuo de Libereco  - koordinatoj 40°41′21″N 74°02′40″W   /   40.68917°N , 74.04444°U  / 40.68917; -74.04444  ( Statuo de Libereco ) Alteco de statuo 46 m  - alteco kun soklo 93 m Pezo 205 000 kg (451 948 lb) Materialoj bronzo, ŝtalo Proponinto Gustave Eiffel  - skulptisto Frédéric Auguste Bartholdi  - proponinto de soklo Richard Morris Hunt  - kapo laŭ modelo de Isabella Eugenie Boyer Transdonita 4-an de julio 1886 Por publiko alirebla por publiko Horzono OT (UTC-5)  - somera tempo OT (UTC-4) Loko
Read more

Tanganjiko

Image
Lago Tanganjiko Situo centra Afriko Ĉefaj fontoj Kalambo, Lufuko, Malagarasi kaj Ruzizi Ĉefaj elfluoj Lukugo Setlejoj Bujumbura, Kalemie Datenoj Koordinatoj 6°  S , 30°  O -6.3827777777778 29.615 782 Koordinatoj: 6°  S , 30°  O DEC Supermara alteco 782 m Surfaca areo 32 893 km² f5 Maksimuma longo 673 km f6 Maksimuma larĝo 72 km f7 Akva volumeno 18.900 km³ dep1 f8 Maksimuma profundo 1470 m f10 Averaĝa profundo 570 m f11 v   •   d   •   r Mapo de la lago Tanganjiko estas lago en orienta Afriko, inter Burundo, Kongo Kinŝasa, Tanzanio kaj Zambio. Ĝi apartenas al la Granda Rifto kaj estas unu el la Afrikaj Grandaj Lagoj. Tanzanio (46%) kaj DRK (40%) havas la majoritaton de la lago. Ĝi havas longon de 673 km, areon de ĉ. 32.900 km² kaj maksimuman profundon taksatan je 1470 metroj; pro tiu ĉi granda profundo ankaŭ ĝia volumeno estas sufiĉe granda, 18.900 km³. Oni konsi
Read more

Liste der Baudenkmäler in Enneberg

Image
Die Liste der Baudenkmäler in Enneberg (ladinisch Mareo , italienisch Marebbe ) enthält die 36 als Baudenkmäler ausgewiesenen Objekte auf dem Gebiet der Gemeinde Enneberg in Südtirol. Basis ist das im Internet einsehbare offizielle Verzeichnis der Baudenkmäler in Südtirol. Dabei kann es sich beispielsweise um Sakralbauten, Wohnhäuser, Bauernhöfe und Adelsansitze handeln. Die Reihenfolge in dieser Liste orientiert sich an der Bezeichnung, alternativ ist sie auch nach der Adresse oder dem Datum der Unterschutzstellung sortierbar. Liste | Foto   Bezeichnung Standort Eintragung Beschreibung Metadaten ja Altes Gasthaus Janesch ID: 14567 46° 42′ 54″ N, 11° 52′ 42″ O 76882557800 0 8. Sep. 1986 (BLR-LAB 4974) Ländliches Wohnhaus/Bauernhof KG:  Welschellen Bauparzelle: 391 ja Asch (Prack) ID: 14540 46° 43′ 12″ N, 11° 54′ 35″ O 75496485300 20. Nov. 1950 (MD) Ansitz KG:  Enneberg Bauparzelle: 38 ja Biccia mit Stade
Read more