Mixing
How do we find out angle from $x$ & $y$ coordinates?
up vote
6
down vote
favorite
I found the following sentence.
To find the angle you use the arctangent function like this, angle $=tan^{-1}left(frac{y}{x}right)$.
But I am curious, is this the only way to know the angle?
In other words, is it possible to find the angle with $sinleft(frac{y}{x}right)$, $cosleft(frac{y}{x}right)$ or $tan$.. etc?
trigonometry arithmetic
edited Sep 18 '17 at 17:46
A---B
2,26611337
asked Jun 16 '15 at 8:51
gmotree
44117
add a comment |
up vote
6
down vote
favorite
I found the following sentence.
To find the angle you use the arctangent function like this, angle $=tan^{-1}left(frac{y}{x}right)$.
But I am curious, is this the only way to know the angle?
In other words, is it possible to find the angle with $sinleft(frac{y}{x}right)$, $cosleft(frac{y}{x}right)$ or $tan$.. etc?
trigonometry arithmetic
edited Sep 18 '17 at 17:46
A---B
2,26611337
asked Jun 16 '15 at 8:51
gmotree
44117
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I found the following sentence.
To find the angle you use the arctangent function like this, angle $=tan^{-1}left(frac{y}{x}right)$.
But I am curious, is this the only way to know the angle?
In other words, is it possible to find the angle with $sinleft(frac{y}{x}right)$, $cosleft(frac{y}{x}right)$ or $tan$.. etc?
trigonometry arithmetic
edited Sep 18 '17 at 17:46
A---B
2,26611337
asked Jun 16 '15 at 8:51
gmotree
44117
I found the following sentence.
To find the angle you use the arctangent function like this, angle $=tan^{-1}left(frac{y}{x}right)$.
But I am curious, is this the only way to know the angle?
In other words, is it possible to find the angle with $sinleft(frac{y}{x}right)$, $cosleft(frac{y}{x}right)$ or $tan$.. etc?
trigonometry arithmetic
trigonometry arithmetic
edited Sep 18 '17 at 17:46
A---B
2,26611337
asked Jun 16 '15 at 8:51
gmotree
44117
edited Sep 18 '17 at 17:46
A---B
2,26611337
asked Jun 16 '15 at 8:51
gmotree
44117
edited Sep 18 '17 at 17:46
A---B
2,26611337
edited Sep 18 '17 at 17:46
A---B
2,26611337
edited Sep 18 '17 at 17:46
A---B
2,26611337
2,26611337
asked Jun 16 '15 at 8:51
gmotree
44117
asked Jun 16 '15 at 8:51
gmotree
44117
asked Jun 16 '15 at 8:51
gmotree
44117
44117
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
For any given point $(x, y)$, the angle say $theta$ of the line, passing through this point & the origin, with the positive x-direction is given as $$color{blue}{tantheta=frac{y}{x}}$$ While other values are given as
$$color{blue}{sintheta=frac{y}{sqrt{x^2+y^2}}}$$
$$color{blue}{costheta=frac{x}{sqrt{x^2+y^2}}}$$
answered Jun 16 '15 at 11:16
Harish Chandra Rajpoot
29.4k103671
How can you find out whether degree or radian?
– gmotree
Jun 16 '15 at 21:45
The angle obtained is generally in radian which can be changed in degree by multiplying by $frac{180^o}{pi}$
– Harish Chandra Rajpoot
Jun 17 '15 at 3:45
Can I get a hyperbolic function like arctanh sinh tanh.. from sine or cosine function?
– gmotree
Jun 17 '15 at 5:07
add a comment |
up vote
4
down vote
Does the picture below help you visualise this? By 'angle' we mean $theta$ below in plane polar coordinates. For some point $(x_0,y_0)$ on the plane, we can solve for $theta$ using trigonometry.
answered Jun 16 '15 at 8:57
Ruvi Lecamwasam
1,250618
Thanks art pictures. Yes so I asking about can we know the angle from any other tri- functions.
– gmotree
Jun 16 '15 at 9:09
This answer would be improved my including the actual trig to get theta from x0 and y0 here.
– lindes
Jun 1 at 4:41
add a comment |
up vote
0
down vote
You can use one of the following three formulas to find an angle.
1)$$f(x,y)=pi-frac{pi}{1+mbox{sgn}(x)}left(1-mbox{sgn}(y^2)right)-frac{pi}{4}left(2+mbox{sgn}(x)right)mbox{sgn}(y)$$
$$-mbox{sgn}(xy)*mbox{atan}left(frac{|x|-|y|}{|x|+|y|}right)$$
2)$$f(x,y)=pi-pi/2*(1+mbox{sgn}(x))(1-mbox{sgn}(y^2))-frac{pi}{4}(2+mbox{sgn}(x))mbox{sgn}(y)$$
$$-mbox{sgn}(xy)mbox{asin}left(frac{left|xright|-left|yright|}{sqrt{2*x^2+2*y^2}}right)$$
3)$$f(x,y)=pi-frac{pi}{2}(1+mbox{sgn}(x))(1-mbox{sgn}(y^2))-frac{pi}{4}(2+mbox{sgn}(x))mbox{sgn}(y)$$
$$-mbox{sgn}(left|xright|-left|yright|)mbox{sgn}(xy)mbox{acos}left(frac{left|xright|+left|yright|}{sqrt{2*x^2+2*y^2}}right)$$
Each of the formulas give the angle from $0$ to $2pi$ for any value of $x$ and $y$.
For $x=y=0$, the result is undefined.
edited 22 hours ago
Alexander Gruber♦
20.1k24102171
answered Nov 16 at 0:11
theodore panagos
211
New contributor
theodore panagos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you try to use MathJax to format the formula; as is this is hard to read. math.meta.stackexchange.com/questions/5020/…
– quid♦
2 days ago
add a comment |
up vote
0
down vote
In this recent answer, it is shown that
$$
theta=2arctanleft(vcenter{frac y{x+sqrt{x^2+y^2}}}right)
$$
This formula works for all $x,y$ except on the negative real axis, where $theta$ goes from just under $pi$ on top to just above $-pi$ underneath.
answered 21 hours ago
robjohn♦
262k27300620
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
For any given point $(x, y)$, the angle say $theta$ of the line, passing through this point & the origin, with the positive x-direction is given as $$color{blue}{tantheta=frac{y}{x}}$$ While other values are given as
$$color{blue}{sintheta=frac{y}{sqrt{x^2+y^2}}}$$
$$color{blue}{costheta=frac{x}{sqrt{x^2+y^2}}}$$
answered Jun 16 '15 at 11:16
Harish Chandra Rajpoot
29.4k103671
How can you find out whether degree or radian?
– gmotree
Jun 16 '15 at 21:45
The angle obtained is generally in radian which can be changed in degree by multiplying by $frac{180^o}{pi}$
– Harish Chandra Rajpoot
Jun 17 '15 at 3:45
Can I get a hyperbolic function like arctanh sinh tanh.. from sine or cosine function?
– gmotree
Jun 17 '15 at 5:07
add a comment |
up vote
4
down vote
accepted
For any given point $(x, y)$, the angle say $theta$ of the line, passing through this point & the origin, with the positive x-direction is given as $$color{blue}{tantheta=frac{y}{x}}$$ While other values are given as
$$color{blue}{sintheta=frac{y}{sqrt{x^2+y^2}}}$$
$$color{blue}{costheta=frac{x}{sqrt{x^2+y^2}}}$$
answered Jun 16 '15 at 11:16
Harish Chandra Rajpoot
29.4k103671
How can you find out whether degree or radian?
– gmotree
Jun 16 '15 at 21:45
The angle obtained is generally in radian which can be changed in degree by multiplying by $frac{180^o}{pi}$
– Harish Chandra Rajpoot
Jun 17 '15 at 3:45
Can I get a hyperbolic function like arctanh sinh tanh.. from sine or cosine function?
– gmotree
Jun 17 '15 at 5:07
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
For any given point $(x, y)$, the angle say $theta$ of the line, passing through this point & the origin, with the positive x-direction is given as $$color{blue}{tantheta=frac{y}{x}}$$ While other values are given as
$$color{blue}{sintheta=frac{y}{sqrt{x^2+y^2}}}$$
$$color{blue}{costheta=frac{x}{sqrt{x^2+y^2}}}$$
answered Jun 16 '15 at 11:16
Harish Chandra Rajpoot
29.4k103671
For any given point $(x, y)$, the angle say $theta$ of the line, passing through this point & the origin, with the positive x-direction is given as $$color{blue}{tantheta=frac{y}{x}}$$ While other values are given as
$$color{blue}{sintheta=frac{y}{sqrt{x^2+y^2}}}$$
$$color{blue}{costheta=frac{x}{sqrt{x^2+y^2}}}$$
answered Jun 16 '15 at 11:16
Harish Chandra Rajpoot
29.4k103671
answered Jun 16 '15 at 11:16
Harish Chandra Rajpoot
29.4k103671
answered Jun 16 '15 at 11:16
Harish Chandra Rajpoot
29.4k103671
answered Jun 16 '15 at 11:16
Harish Chandra Rajpoot
29.4k103671
29.4k103671
How can you find out whether degree or radian?
– gmotree
Jun 16 '15 at 21:45
The angle obtained is generally in radian which can be changed in degree by multiplying by $frac{180^o}{pi}$
– Harish Chandra Rajpoot
Jun 17 '15 at 3:45
Can I get a hyperbolic function like arctanh sinh tanh.. from sine or cosine function?
– gmotree
Jun 17 '15 at 5:07
add a comment |
How can you find out whether degree or radian?
– gmotree
Jun 16 '15 at 21:45
The angle obtained is generally in radian which can be changed in degree by multiplying by $frac{180^o}{pi}$
– Harish Chandra Rajpoot
Jun 17 '15 at 3:45
Can I get a hyperbolic function like arctanh sinh tanh.. from sine or cosine function?
– gmotree
Jun 17 '15 at 5:07
How can you find out whether degree or radian?
– gmotree
Jun 16 '15 at 21:45
How can you find out whether degree or radian?
– gmotree
Jun 16 '15 at 21:45
The angle obtained is generally in radian which can be changed in degree by multiplying by $frac{180^o}{pi}$
– Harish Chandra Rajpoot
Jun 17 '15 at 3:45
The angle obtained is generally in radian which can be changed in degree by multiplying by $frac{180^o}{pi}$
– Harish Chandra Rajpoot
Jun 17 '15 at 3:45
Can I get a hyperbolic function like arctanh sinh tanh.. from sine or cosine function?
– gmotree
Jun 17 '15 at 5:07
Can I get a hyperbolic function like arctanh sinh tanh.. from sine or cosine function?
– gmotree
Jun 17 '15 at 5:07
add a comment |
up vote
4
down vote
Does the picture below help you visualise this? By 'angle' we mean $theta$ below in plane polar coordinates. For some point $(x_0,y_0)$ on the plane, we can solve for $theta$ using trigonometry.
answered Jun 16 '15 at 8:57
Ruvi Lecamwasam
1,250618
Thanks art pictures. Yes so I asking about can we know the angle from any other tri- functions.
– gmotree
Jun 16 '15 at 9:09
This answer would be improved my including the actual trig to get theta from x0 and y0 here.
– lindes
Jun 1 at 4:41
add a comment |
up vote
4
down vote
Does the picture below help you visualise this? By 'angle' we mean $theta$ below in plane polar coordinates. For some point $(x_0,y_0)$ on the plane, we can solve for $theta$ using trigonometry.
answered Jun 16 '15 at 8:57
Ruvi Lecamwasam
1,250618
Thanks art pictures. Yes so I asking about can we know the angle from any other tri- functions.
– gmotree
Jun 16 '15 at 9:09
This answer would be improved my including the actual trig to get theta from x0 and y0 here.
– lindes
Jun 1 at 4:41
add a comment |
up vote
4
down vote
up vote
4
down vote
Does the picture below help you visualise this? By 'angle' we mean $theta$ below in plane polar coordinates. For some point $(x_0,y_0)$ on the plane, we can solve for $theta$ using trigonometry.
answered Jun 16 '15 at 8:57
Ruvi Lecamwasam
1,250618
Does the picture below help you visualise this? By 'angle' we mean $theta$ below in plane polar coordinates. For some point $(x_0,y_0)$ on the plane, we can solve for $theta$ using trigonometry.
answered Jun 16 '15 at 8:57
Ruvi Lecamwasam
1,250618
answered Jun 16 '15 at 8:57
Ruvi Lecamwasam
1,250618
answered Jun 16 '15 at 8:57
Ruvi Lecamwasam
1,250618
answered Jun 16 '15 at 8:57
Ruvi Lecamwasam
1,250618
1,250618
Thanks art pictures. Yes so I asking about can we know the angle from any other tri- functions.
– gmotree
Jun 16 '15 at 9:09
This answer would be improved my including the actual trig to get theta from x0 and y0 here.
– lindes
Jun 1 at 4:41
add a comment |
Thanks art pictures. Yes so I asking about can we know the angle from any other tri- functions.
– gmotree
Jun 16 '15 at 9:09
This answer would be improved my including the actual trig to get theta from x0 and y0 here.
– lindes
Jun 1 at 4:41
Thanks art pictures. Yes so I asking about can we know the angle from any other tri- functions.
– gmotree
Jun 16 '15 at 9:09
Thanks art pictures. Yes so I asking about can we know the angle from any other tri- functions.
– gmotree
Jun 16 '15 at 9:09
This answer would be improved my including the actual trig to get theta from x0 and y0 here.
– lindes
Jun 1 at 4:41
This answer would be improved my including the actual trig to get theta from x0 and y0 here.
– lindes
Jun 1 at 4:41
add a comment |
up vote
0
down vote
You can use one of the following three formulas to find an angle.
1)$$f(x,y)=pi-frac{pi}{1+mbox{sgn}(x)}left(1-mbox{sgn}(y^2)right)-frac{pi}{4}left(2+mbox{sgn}(x)right)mbox{sgn}(y)$$
$$-mbox{sgn}(xy)*mbox{atan}left(frac{|x|-|y|}{|x|+|y|}right)$$
2)$$f(x,y)=pi-pi/2*(1+mbox{sgn}(x))(1-mbox{sgn}(y^2))-frac{pi}{4}(2+mbox{sgn}(x))mbox{sgn}(y)$$
$$-mbox{sgn}(xy)mbox{asin}left(frac{left|xright|-left|yright|}{sqrt{2*x^2+2*y^2}}right)$$
3)$$f(x,y)=pi-frac{pi}{2}(1+mbox{sgn}(x))(1-mbox{sgn}(y^2))-frac{pi}{4}(2+mbox{sgn}(x))mbox{sgn}(y)$$
$$-mbox{sgn}(left|xright|-left|yright|)mbox{sgn}(xy)mbox{acos}left(frac{left|xright|+left|yright|}{sqrt{2*x^2+2*y^2}}right)$$
Each of the formulas give the angle from $0$ to $2pi$ for any value of $x$ and $y$.
For $x=y=0$, the result is undefined.
edited 22 hours ago
Alexander Gruber♦
20.1k24102171
answered Nov 16 at 0:11
theodore panagos
211
New contributor
theodore panagos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you try to use MathJax to format the formula; as is this is hard to read. math.meta.stackexchange.com/questions/5020/…
– quid♦
2 days ago
add a comment |
up vote
0
down vote
You can use one of the following three formulas to find an angle.
1)$$f(x,y)=pi-frac{pi}{1+mbox{sgn}(x)}left(1-mbox{sgn}(y^2)right)-frac{pi}{4}left(2+mbox{sgn}(x)right)mbox{sgn}(y)$$
$$-mbox{sgn}(xy)*mbox{atan}left(frac{|x|-|y|}{|x|+|y|}right)$$
2)$$f(x,y)=pi-pi/2*(1+mbox{sgn}(x))(1-mbox{sgn}(y^2))-frac{pi}{4}(2+mbox{sgn}(x))mbox{sgn}(y)$$
$$-mbox{sgn}(xy)mbox{asin}left(frac{left|xright|-left|yright|}{sqrt{2*x^2+2*y^2}}right)$$
3)$$f(x,y)=pi-frac{pi}{2}(1+mbox{sgn}(x))(1-mbox{sgn}(y^2))-frac{pi}{4}(2+mbox{sgn}(x))mbox{sgn}(y)$$
$$-mbox{sgn}(left|xright|-left|yright|)mbox{sgn}(xy)mbox{acos}left(frac{left|xright|+left|yright|}{sqrt{2*x^2+2*y^2}}right)$$
Each of the formulas give the angle from $0$ to $2pi$ for any value of $x$ and $y$.
For $x=y=0$, the result is undefined.
edited 22 hours ago
Alexander Gruber♦
20.1k24102171
answered Nov 16 at 0:11
theodore panagos
211
New contributor
theodore panagos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you try to use MathJax to format the formula; as is this is hard to read. math.meta.stackexchange.com/questions/5020/…
– quid♦
2 days ago
add a comment |
up vote
0
down vote
up vote
0
down vote
You can use one of the following three formulas to find an angle.
1)$$f(x,y)=pi-frac{pi}{1+mbox{sgn}(x)}left(1-mbox{sgn}(y^2)right)-frac{pi}{4}left(2+mbox{sgn}(x)right)mbox{sgn}(y)$$
$$-mbox{sgn}(xy)*mbox{atan}left(frac{|x|-|y|}{|x|+|y|}right)$$
2)$$f(x,y)=pi-pi/2*(1+mbox{sgn}(x))(1-mbox{sgn}(y^2))-frac{pi}{4}(2+mbox{sgn}(x))mbox{sgn}(y)$$
$$-mbox{sgn}(xy)mbox{asin}left(frac{left|xright|-left|yright|}{sqrt{2*x^2+2*y^2}}right)$$
3)$$f(x,y)=pi-frac{pi}{2}(1+mbox{sgn}(x))(1-mbox{sgn}(y^2))-frac{pi}{4}(2+mbox{sgn}(x))mbox{sgn}(y)$$
$$-mbox{sgn}(left|xright|-left|yright|)mbox{sgn}(xy)mbox{acos}left(frac{left|xright|+left|yright|}{sqrt{2*x^2+2*y^2}}right)$$
Each of the formulas give the angle from $0$ to $2pi$ for any value of $x$ and $y$.
For $x=y=0$, the result is undefined.
edited 22 hours ago
Alexander Gruber♦
20.1k24102171
answered Nov 16 at 0:11
theodore panagos
211
New contributor
theodore panagos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You can use one of the following three formulas to find an angle.
1)$$f(x,y)=pi-frac{pi}{1+mbox{sgn}(x)}left(1-mbox{sgn}(y^2)right)-frac{pi}{4}left(2+mbox{sgn}(x)right)mbox{sgn}(y)$$
$$-mbox{sgn}(xy)*mbox{atan}left(frac{|x|-|y|}{|x|+|y|}right)$$
2)$$f(x,y)=pi-pi/2*(1+mbox{sgn}(x))(1-mbox{sgn}(y^2))-frac{pi}{4}(2+mbox{sgn}(x))mbox{sgn}(y)$$
$$-mbox{sgn}(xy)mbox{asin}left(frac{left|xright|-left|yright|}{sqrt{2*x^2+2*y^2}}right)$$
3)$$f(x,y)=pi-frac{pi}{2}(1+mbox{sgn}(x))(1-mbox{sgn}(y^2))-frac{pi}{4}(2+mbox{sgn}(x))mbox{sgn}(y)$$
$$-mbox{sgn}(left|xright|-left|yright|)mbox{sgn}(xy)mbox{acos}left(frac{left|xright|+left|yright|}{sqrt{2*x^2+2*y^2}}right)$$
Each of the formulas give the angle from $0$ to $2pi$ for any value of $x$ and $y$.
For $x=y=0$, the result is undefined.
edited 22 hours ago
Alexander Gruber♦
20.1k24102171
answered Nov 16 at 0:11
theodore panagos
211
New contributor
theodore panagos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 22 hours ago
Alexander Gruber♦
20.1k24102171
edited 22 hours ago
Alexander Gruber♦
20.1k24102171
edited 22 hours ago
Alexander Gruber♦
20.1k24102171
20.1k24102171
answered Nov 16 at 0:11
theodore panagos
211
New contributor
theodore panagos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 16 at 0:11
theodore panagos
211
answered Nov 16 at 0:11
theodore panagos
211
211
New contributor
theodore panagos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
theodore panagos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
theodore panagos is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you try to use MathJax to format the formula; as is this is hard to read. math.meta.stackexchange.com/questions/5020/…
– quid♦
2 days ago
add a comment |
Could you try to use MathJax to format the formula; as is this is hard to read. math.meta.stackexchange.com/questions/5020/…
– quid♦
2 days ago
Could you try to use MathJax to format the formula; as is this is hard to read. math.meta.stackexchange.com/questions/5020/…
– quid♦
2 days ago
Could you try to use MathJax to format the formula; as is this is hard to read. math.meta.stackexchange.com/questions/5020/…
– quid♦
2 days ago
add a comment |
up vote
0
down vote
In this recent answer, it is shown that
$$
theta=2arctanleft(vcenter{frac y{x+sqrt{x^2+y^2}}}right)
$$
This formula works for all $x,y$ except on the negative real axis, where $theta$ goes from just under $pi$ on top to just above $-pi$ underneath.
answered 21 hours ago
robjohn♦
262k27300620
add a comment |
up vote
0
down vote
In this recent answer, it is shown that
$$
theta=2arctanleft(vcenter{frac y{x+sqrt{x^2+y^2}}}right)
$$
This formula works for all $x,y$ except on the negative real axis, where $theta$ goes from just under $pi$ on top to just above $-pi$ underneath.
answered 21 hours ago
robjohn♦
262k27300620
add a comment |
up vote
0
down vote
up vote
0
down vote
In this recent answer, it is shown that
$$
theta=2arctanleft(vcenter{frac y{x+sqrt{x^2+y^2}}}right)
$$
This formula works for all $x,y$ except on the negative real axis, where $theta$ goes from just under $pi$ on top to just above $-pi$ underneath.
answered 21 hours ago
robjohn♦
262k27300620
In this recent answer, it is shown that
$$
theta=2arctanleft(vcenter{frac y{x+sqrt{x^2+y^2}}}right)
$$
This formula works for all $x,y$ except on the negative real axis, where $theta$ goes from just under $pi$ on top to just above $-pi$ underneath.
answered 21 hours ago
robjohn♦
262k27300620
answered 21 hours ago
robjohn♦
262k27300620
answered 21 hours ago
robjohn♦
262k27300620
answered 21 hours ago
robjohn♦
262k27300620
262k27300620
add a comment |
add a comment |
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