# How do you convert (-1/2, -sqrt3/2) into polar form?

Nov 18, 2016

$\left(1 , \frac{4 \pi}{3}\right)$

#### Explanation:

To convert from $\textcolor{b l u e}{\text{cartesian to polar form}}$

That is $\left(x , y\right) \to \left(r , \theta\right)$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{r = \sqrt{{x}^{2} + {y}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

and $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

Here $x = - \frac{1}{2} \text{ and } y = - \frac{\sqrt{3}}{2}$

$\Rightarrow r = \sqrt{{\left(- \frac{1}{2}\right)}^{2} + {\left(- \frac{\sqrt{3}}{2}\right)}^{2}} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$

now $\left(- \frac{1}{2} , - \frac{\sqrt{3}}{2}\right)$ is in the 3rd quadrant so we must ensure that $\theta$ is in the 3rd quadrant.

$\Rightarrow \theta = {\tan}^{-} 1 \left(\frac{- \frac{\sqrt{3}}{2}}{- \frac{1}{2}}\right)$

$= {\tan}^{-} 1 \left(\sqrt{3}\right) = \frac{\pi}{3} \leftarrow \text{ reference angle}$

$\Rightarrow \theta = \left(\pi + \frac{\pi}{3}\right) = \frac{4 \pi}{3} \leftarrow \text{ 3rd quadrant}$

$\Rightarrow \left(- \frac{1}{2} , - \frac{\sqrt{3}}{2}\right) \to \left(1 , \frac{4 \pi}{3}\right)$

Nov 18, 2016

#### Explanation:

The polar coordinate system is an ordered pair, $\left(r , \theta\right)$.

To compute r from from Cartesian coordinates, $\left(x , y\right)$, use the equation:

$r = \sqrt{{x}^{2} + {y}^{2}}$

$r = \sqrt{{\left(- \frac{1}{2}\right)}^{2} + {\left(- \frac{\sqrt{3}}{2}\right)}^{2}}$

r = sqrt((1/4 + 3/4)

$r = 1$

To compute $\theta$ from Cartesian coordinates, $\left(x , y\right)$, use the appropriate one of the following equations:

1. If $x > 0 \mathmr{and} y \ge 0$m then use: $\theta = {\tan}^{-} 1 \left(\frac{y}{x}\right)$
2. If $x = 0 \mathmr{and} y > 0$, then use: $\theta = \frac{\pi}{2}$
3. If $x = 0 \mathmr{and} y < 0$, then use: $\theta = \frac{3 \pi}{2}$
4. If $x < 0$, then use: $\theta = \pi + {\tan}^{-} 1 \left(\frac{y}{x}\right)$
5. If $x > 0 \mathmr{and} y < 0$, then use: $\theta = 2 \pi + {\tan}^{-} 1 \left(\frac{y}{x}\right)$

The appropriate one is equation 4:

$\theta = \pi + {\tan}^{-} 1 \left(\frac{- \frac{\sqrt{3}}{2}}{- \frac{1}{2}}\right)$

$\theta = \pi + {\tan}^{-} 1 \left(\sqrt{3}\right)$

$\theta = \pi + \frac{\pi}{3}$

$\theta = \frac{4 \pi}{3}$

The polar point is $\left(1 , \frac{4 \pi}{3}\right)$