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calculator's angle answer for trig ratios that can work in more than 1 quadrant on the unit circle
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraUnderstanding inverse trigonometric relationsFinding a point on the unit circle; more specifically, what quadrant it is inBroken Calculator: only certain unary functions work.How does the unit circle work for trigonometric ratios of non-acute angles?unit circle trigonometry where angle is greater than 90 degrees.Why are the Trig functions defined by the counterclockwise path of a circle?Trigonometric Ratios for angles greater than 90 degrees and the Unit CircleIf $sinx=t, quad xin(frac3pi2,2pi),$ what is $tanx?$Trigonometric Ratios for angles greater than 90 degrees in unit circleHow does the unit circle work for trigonometric ratios of obtuse angles?Why we need an angle to for trig ratios?
$begingroup$
Why does the calculator do a cc (counterclockwise) rotation for positive trig ratios instead of clockwise,
and a clockwise rotation for negative sine & tan instead of cc
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
edited 2 days ago
N. F. Taussig
45.5k103358
asked 2 days ago
Allan HenriquesAllan Henriques
688
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.
$endgroup$
add a comment |
$begingroup$
Why does the calculator do a cc (counterclockwise) rotation for positive trig ratios instead of clockwise,
and a clockwise rotation for negative sine & tan instead of cc
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
edited 2 days ago
N. F. Taussig
45.5k103358
asked 2 days ago
Allan HenriquesAllan Henriques
688
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 days 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 days ago
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
2 days ago
add a comment |
$begingroup$
Why does the calculator do a cc (counterclockwise) rotation for positive trig ratios instead of clockwise,
and a clockwise rotation for negative sine & tan instead of cc
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
edited 2 days ago
N. F. Taussig
45.5k103358
asked 2 days ago
Allan HenriquesAllan Henriques
688
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.
$endgroup$
Why does the calculator do a cc (counterclockwise) rotation for positive trig ratios instead of clockwise,
and a clockwise rotation for negative sine & tan instead of cc
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
edited 2 days ago
N. F. Taussig
45.5k103358
asked 2 days ago
Allan HenriquesAllan Henriques
688
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 2 days ago
N. F. Taussig
45.5k103358
asked 2 days ago
Allan HenriquesAllan Henriques
688
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 2 days ago
N. F. Taussig
45.5k103358
edited 2 days ago
N. F. Taussig
45.5k103358
edited 2 days ago
N. F. Taussig
45.5k103358
45.5k103358
asked 2 days ago
Allan HenriquesAllan Henriques
688
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 days ago
Allan HenriquesAllan Henriques
688
asked 2 days ago
Allan HenriquesAllan Henriques
688
688
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 days 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 days ago
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
2 days ago
add a comment |
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 days 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 days ago
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
2 days 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 days 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 days 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 days 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 days ago
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
2 days ago
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered 2 days ago
DMcMorDMcMor
3,11521431
$endgroup$
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 days ago
add a comment |
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$begingroup$
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.
answered 2 days ago
DMcMorDMcMor
3,11521431
$endgroup$
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 days 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.
answered 2 days ago
DMcMorDMcMor
3,11521431
$endgroup$
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 days 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.
answered 2 days ago
DMcMorDMcMor
3,11521431
$endgroup$
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 $-fracpi2leq x leq fracpi2$, 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.
answered 2 days ago
DMcMorDMcMor
3,11521431
answered 2 days ago
DMcMorDMcMor
3,11521431
answered 2 days ago
DMcMorDMcMor
3,11521431
answered 2 days ago
DMcMorDMcMor
3,11521431
3,11521431
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 days ago
add a comment |
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 days 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 days 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 days ago
add a comment |
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.
Allan Henriques is a new contributor. Be nice, and check out our Code of Conduct.
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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 days 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 days ago
$begingroup$
This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
2 days ago