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

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

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 $-frac{pi}{2} leq x leq frac{pi}{2}$, 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 | 

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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:

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

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

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 $-frac{pi}{2} leq x leq frac{pi}{2}$, 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