What is the theta in this cartesian to polar coordinates problem?

If x=1 and y=1, what is the theta and why?

I know for a fact that the answer is pi/4 but I do not get why.

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You mean to ask why is the angle $\theta$ such that $(\cos (\theta), \sin (\theta))=(x,y)=(1,1)$ equal to $\pi /4$? – Git Gud Sep 30 '13 at 21:56
I know that its equivalent polar coordiantes is (squareroot of 2, pi/4), but I do not get why. – Gannicus Sep 30 '13 at 21:58

Given $(x,y)\in \Bbb R^2$, you can find $\rho\in \Bbb R^+$ and $\theta \in [-\pi,\pi]$ such that $(x,y)=\rho (\cos (\theta), \sin (\theta))$, where $\rho=\sqrt {x^2+y^2}$ and, with $\tan$ defined on $\left[-\pi , -\frac \pi 2\right[\cup \left]-\frac \pi 2, 0\right]$ (so that $\theta \in [-\pi,\pi]$), $$\theta =\begin{cases} \arctan \left(\dfrac yx\right), &\text{if } x>0\\ \dfrac \pi 2, &\text{if }x=0 \text{ and }y>0\\ \arctan \left(\dfrac yx\right)+\pi &\text{if }x<0\\ -\dfrac \pi 2, &\text{if } x=0\text{ and }y<0\end{cases}$$

In your example you get $\rho=\sqrt {1^2+1^2}$ and $\theta=\arctan \left(\dfrac 11\right)=\arctan \left( 1\right)=\dfrac \pi 4.$

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@labbhattacharjee Nice addition. Thanks. – Git Gud Nov 10 '13 at 14:30

The angle $\theta$ in polar coordinates is measured from the $+x$ axis, where $\theta=0$ increasing counterclockwise. Since the vector $(1,1)$ is rotated $\frac \pi 4$ from the $+x$ axis, that is the $\theta$ coordinate of the point. In general, the $\theta$ coordinate $(x,y)$ is $\arctan \frac yx (+ \pi)$ where the $(+ \pi)$ indicates that there are two angles along the line which differ by $\pi$ and you need to figure out which you want. If the point were $(-1,1)$, for example, we have $\frac yx=1$, but the angle is $\frac {5\pi }4$

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If the point were (-1,1), can I use 7π/4 instead? – Gannicus Sep 30 '13 at 22:01
@Gannicus No, you'd use $\dfrac{3\pi}{4}$, as the point is in the $2$nd quadrant. – amWhy Sep 30 '13 at 22:03
Ah, my bad. But can I use 5π/6? That is also in the 2nd quadrant. – Gannicus Sep 30 '13 at 22:04
@Gannicus: That is the $\pi$ uncertainty I was talking about. – Ross Millikan Sep 30 '13 at 22:05
@Gannicus: but $\arctan -1 \neq \frac {5\pi }6$ – Ross Millikan Sep 30 '13 at 22:06

$$x = r\cos \theta = 1\;\text{ and }\;y = r\sin\theta = 1$$ $$\tan\theta = \dfrac {\sin\theta}{\cos \theta} = \dfrac xy = 1$$

$$\theta = \arctan 1 = \dfrac \pi 4$$

$r$ is simply $\sqrt{1^2 + 1^2 } = \sqrt 2$. Recall, the radius is equivalent to $r$ where $$x^2 + y^2 = r^2$$

So $(x, y) = (1, 1)$ in Cartesian Coordinates is equivalent to $(r, \theta) = (\sqrt 2, \pi 4)$ in polar coordinates.

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