# How do we find out angle from $x$ & $y$ coordinates?

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 $\sin\left(\frac{y}{x}\right)$, $\cos\left(\frac{y}{x}\right)$ or $\tan$.. etc?

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 $$\color{blue}{\tan\theta=\frac{y}{x}}$$ While other values are given as
$$\color{blue}{\sin\theta=\frac{y}{\sqrt{x^2+y^2}}}$$ $$\color{blue}{\cos\theta=\frac{x}{\sqrt{x^2+y^2}}}$$

• How can you find out whether degree or radian? Jun 16, 2015 at 21:45
• The angle obtained is generally in radian which can be changed in degree by multiplying by $\frac{180^o}{\pi}$ Jun 17, 2015 at 3:45
• Can I get a hyperbolic function like arctanh sinh tanh.. from sine or cosine function? Jun 17, 2015 at 5:07

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.

• Thanks art pictures. Yes so I asking about can we know the angle from any other tri- functions. Jun 16, 2015 at 9:09
• This answer would be improved my including the actual trig to get theta from x0 and y0 here. Jun 1, 2018 at 4:41

In this recent answer, it is shown that $$\theta=2\arctan\left(\vcenter{\frac y{x+\sqrt{x^2+y^2}}}\right)$$ 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.

• Thank you so much! You saved my day. Dec 6, 2018 at 11:10

You can use one of the following three formulas to find an angle.

1)$$f(x,y)=\pi-\frac{\pi}{2}(1+\mbox{sgn}(x))(1-\mbox{sgn}(y^2))-\frac{\pi}{4}\left(2+\mbox{sgn}(x)\right)\mbox{sgn}(y)$$

$$-\mbox{sgn}(xy)*\mbox{atan}\left(\frac{|x|-|y|}{|x|+|y|}\right)$$

2)$$f(x,y)=\pi-\frac{\pi}{2}(1+\mbox{sgn}(x))(1-\mbox{sgn}(y^2))-\frac{\pi}{4}(2+\mbox{sgn}(x))\mbox{sgn}(y)$$

$$-\mbox{sgn}(xy)\mbox{asin}\left(\frac{\left|x\right|-\left|y\right|}{\sqrt{2*x^2+2*y^2}}\right)$$

3)$$f(x,y)=\pi-\frac{\pi}{2}(1+\mbox{sgn}(x))(1-\mbox{sgn}(y^2))-\frac{\pi}{4}(2+\mbox{sgn}(x))\mbox{sgn}(y)$$

$$-\mbox{sgn}(\left|x\right|-\left|y\right|)\mbox{sgn}(xy)\mbox{acos}\left(\frac{\left|x\right|+\left|y\right|}{\sqrt{2*x^2+2*y^2}}\right)$$

Each of the formulas give the angle from $$0$$ to $$2\pi$$ for any value of $$x$$ and $$y$$.

For $$x=y=0$$, the result is undefined.

• Could you try to use MathJax to format the formula; as is this is hard to read. math.meta.stackexchange.com/questions/5020/…
– quid
Nov 17, 2018 at 17:51
• The second term of the first formula must be corrected to be the same Dec 27, 2018 at 15:51
• You can edit your post. I'm not quite sure what you want to change.
– quid
Dec 27, 2018 at 15:53
• The second term of the first formula must be corrected to be the same to the second term of the second and third formula. Dec 27, 2018 at 15:56