# How do you graph the polar equation 1=rcos(theta-pi/6)?

Mar 19, 2018

To graph this easily, we can convert it to rectangular form.

In order to convert, we need to the cosine angle difference formula:

$\cos \left(\textcolor{red}{A} - \textcolor{b l u e}{B}\right) = \cos \textcolor{red}{A} \cos \textcolor{b l u e}{B} + \sin \textcolor{red}{A} \sin \textcolor{b l u e}{B}$

Knowing that $r \sin \theta = y$ and $r \cos \theta = x$, we can convert:

$1 = r \cos \left(\theta - \frac{\pi}{6}\right)$

$1 = r \left(\cos \theta \cos \textcolor{b l a c k}{\frac{\pi}{6}} + \sin \theta \sin \textcolor{b l a c k}{\frac{\pi}{6}}\right)$

$1 = r \left(\cos \theta \cdot \frac{\sqrt{3}}{2} + \sin \theta \cdot \frac{1}{2}\right)$

$1 = r \cos \theta \cdot \frac{\sqrt{3}}{2} + r \sin \theta \cdot \frac{1}{2}$

$1 = x \cdot \frac{\sqrt{3}}{2} + y \cdot \frac{1}{2}$

$1 - x \cdot \frac{\sqrt{3}}{2} = y \cdot \frac{1}{2}$

$2 - x \cdot \sqrt{3} = y$

$y = 2 - x \cdot \sqrt{3}$

$y = - \sqrt{3} x + 2$

Now we can graph this linear equation like any other line.

An easy strategy would be to solve for the $x$- and $y$-intercepts, then connect the dots.

The $x$-intecept occurs when $y = 0$, so:

$\textcolor{w h i t e}{\implies} y = - \sqrt{3} x + 2$

$\implies 0 = - \sqrt{3} x + 2$

$\textcolor{w h i t e}{\implies} \sqrt{3} x = 2$

$\textcolor{w h i t e}{\implies} x = \frac{2}{\sqrt{3}}$

$\textcolor{w h i t e}{\implies} x = \frac{2 \sqrt{3}}{3}$

This means that the $x$-intercept is at $\left(\frac{2 \sqrt{3}}{3} , 0\right)$. Call this point $A$. The $y$-intercept occurs when $x = 0$, so:

$\textcolor{w h i t e}{\implies} y = - \sqrt{3} x + 2$

$\implies y = - \sqrt{3} \cdot 0 + 2$

$\textcolor{w h i t e}{\implies} y = 2$

This means that the $y$-intercept occurs at $\left(0 , 2\right)$. Call this point $B$. Now that we have our two points, we can graph the line:

That's it. Hope this helped!