How do you graph the polar equation $1 = r cos (θ - \frac{π}{6})$ $1=rcos(theta-pi/6)$ ?

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How do you graph the polar equation $1 = r cos (θ - \frac{π}{6})$ $1=rcos(theta-pi/6)$ ?

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Answer 1 · 2018-03-19T05:55:52

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

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

$cos (A - B) = cos A cos B + sin A sin B$ $cos(color(red)A-color(blue)B)=coscolor(red)Acoscolor(blue)B+sincolor(red)Asincolor(blue)B$

Knowing that $r sin θ = y$ $rsintheta=y$ and $r cos θ = x$ $rcostheta=x$ , we can convert:

$1 = r cos (θ - \frac{π}{6})$ $1=rcos(theta-pi/6)$

$1 = r (cos θ cos \frac{π}{6} + sin θ sin \frac{π}{6})$ $1=r(costhetacoscolor(black)(pi/6)+sinthetasincolor(black)(pi/6))$

$1 = r (cos θ \cdot \frac{\sqrt{3}}{2} + sin θ \cdot \frac{1}{2})$ $1=r(costheta*sqrt3/2+sintheta*1/2)$

$1 = r cos θ \cdot \frac{\sqrt{3}}{2} + r sin θ \cdot \frac{1}{2}$ $1=rcostheta*sqrt3/2+rsintheta*1/2$

$1 = x \cdot \frac{\sqrt{3}}{2} + y \cdot \frac{1}{2}$ $1=x*sqrt3/2+y*1/2$

$1 - x \cdot \frac{\sqrt{3}}{2} = y \cdot \frac{1}{2}$ $1-x*sqrt3/2=y*1/2$

$2 - x \cdot \sqrt{3} = y$ $2-x*sqrt3=y$

$y = 2 - x \cdot \sqrt{3}$ $y=2-x*sqrt3$

$y = - \sqrt{3} x + 2$ $y=-sqrt3 x+2$

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

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

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

$\Rightarrow y = - \sqrt{3} x + 2$ $color(white)=>y=-sqrt3 x+2$

$\Rightarrow 0 = - \sqrt{3} x + 2$ $=>0=-sqrt3 x+2$

$\Rightarrow \sqrt{3} x = 2$ $color(white)=>sqrt3 x=2$

$\Rightarrow x = \frac{2}{\sqrt{3}}$ $color(white)=>x=2/sqrt3$

$\Rightarrow x = \frac{2 \sqrt{3}}{3}$ $color(white)=>x=(2sqrt3)/3$

This means that the $x$ $x$ -intercept is at $(\frac{2 \sqrt{3}}{3}, 0)$ $((2sqrt3)/3,0)$ . Call this point $A$ $A$ . The $y$ $y$ -intercept occurs when $x = 0$ $x=0$ , so:

$\Rightarrow y = - \sqrt{3} x + 2$ $color(white)=>y=-sqrt3 x+2$

$\Rightarrow y = - \sqrt{3} \cdot 0 + 2$ $=>y=-sqrt3 *0+2$

$\Rightarrow y = 2$ $color(white)=>y=2$

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

![https://www.desmos.com/calculator](useruploads.socratic.org)

That's it. Hope this helped!

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How do you graph the polar equation $1 = r cos (θ - \frac{π}{6})$ $1=rcos(theta-pi/6)$ ?

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