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?
$\endgroup$4 Answers
$\begingroup$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
$$\color{blue}{\sin\theta=\frac{y}{\sqrt{x^2+y^2}}}$$
$$\color{blue}{\cos\theta=\frac{x}{\sqrt{x^2+y^2}}}$$
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.
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.
$\endgroup$ 1 $\begingroup$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.
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