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












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1 Answer
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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.










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