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# Polar to Rectangular Conversions

## Convert from polar to cartesian coordinates

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Using r and theta to find a Point in the Coordinate Plane

Trig Riddle: I am a point. My Polar coordinates are \begin{align*}(2, 330^\circ)\end{align*}. What are my Cartesian coordinates?

### Convert Polar Coordinates to Cartesian Coordinates

Let's look at how to convert Polar coordinates to Cartesian coordinates. This is, essentially, the reverse of the process used to convert Cartesian coordinates to Polar coordinates.

Given the point \begin{align*}(6, 120^\circ)\end{align*}, let's find the equivalent Cartesian coordinates.

First, consider the diagram below and the right triangle formed by a perpendicular segment to the \begin{align*}x\end{align*}-axis and hypotenuse equal to the radius. We can find the legs of the right triangle using right triangle trigonometry and thus the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} coordinates of the point.

From the diagram we can see that the reference angle is \begin{align*}60^\circ\end{align*}. Now we can use right triangle trigonometry to find \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. In this particular case, we can also use special right triangle ratios or the unit circle.

\begin{align*}\cos 60^\circ &=\frac{x}{6} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \sin 60^\circ =\frac{y}{6} \\ x &=6 \cos 60^\circ=6 \left(\frac{1}{2} \right)=3 \qquad and \qquad \ \ \qquad y =6 \cos 60^\circ =6 \left(\frac{\sqrt{3}}{2} \right)=3\sqrt{3}\end{align*}

Since the point is in the second quadrant, the \begin{align*}x\end{align*} value should be negative giving the Cartesian coordinates \begin{align*}\left(-3, 3 \sqrt{3}\right)\end{align*}.

Recall that every point on the unit circle was \begin{align*}(\cos \theta, \sin \theta)\end{align*}, where \begin{align*}\theta\end{align*} represented the angle of rotation from the positive \begin{align*}x\end{align*} axis and the radius (distance from the origin) was 1. In these problems, our radius varies as we are no longer restricted to the unit circle. In the previous problem, observe that the coordinates \begin{align*}(x, y)\end{align*} are essentially \begin{align*}(6 \cos 60^\circ, 6 \sin 60^\circ)\end{align*} where 6 was the radius and \begin{align*}60^\circ\end{align*} was the reference angle. We could have used the angle of rotation, \begin{align*}120^\circ\end{align*}, and the only difference would be that the cosine ratio would be negative which would automatically make the \begin{align*}x\end{align*} coordinate negative. We can generalize this into a rule for converting from Polar coordinates to Cartesian coordinates:

\begin{align*}(r, \theta)=(r \cos \theta, r \sin \theta)\end{align*}

Now, given the point, \begin{align*}(10, -220^\circ)\end{align*}, let's find the Cartesian coordinates.

Using the rule with \begin{align*}r=10\end{align*} and \begin{align*}\theta=220^\circ\end{align*} and the calculator:

\begin{align*}(10 \cos(-220^\circ), 10 \sin(-220^\circ))=(-7.66, 6.43)\end{align*}

Finally, given the point, \begin{align*}\left(9, \frac{11 \pi}{6} \right)\end{align*}, let's find the exact value of the Cartesian coordinates.

This time \begin{align*}r=9\end{align*} and \begin{align*}\theta=\frac{11 \pi}{6}\end{align*}. So, \begin{align*}\left( 9 \cos \frac{11 \pi}{6}, 9 \sin \frac{11 \pi}{6} \right)=\left(9 \left(\frac{\sqrt{3}}{2} \right), 9 \left(-\frac{1}{2} \right) \right)=\left(\frac{9 \sqrt{3}}{2}, - \frac{9}{2} \right)\end{align*}.

First, draw a diagram. From this diagram we can see that the reference angle is \begin{align*}30^\circ\end{align*}. Now we can use right triangle trigonometry to find \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. In this particular case, we can also use special right triangle ratios or the unit circle.

\begin{align*}\cos 30^\circ &=\frac{x}{2} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \sin 30^\circ =\frac{y}{2} \\ x &=2 \cos 30^\circ=2 \left(\frac{\sqrt{3}}{2} \right)=\sqrt{3} \qquad and \qquad \ \ \qquad y =2 \sin 30^\circ =2 \left(\frac{1}{2} \right)=1\end{align*}

Since the point is in the fourth quadrant, the \begin{align*}y\end{align*} value should be negative giving the Cartesian coordinates \begin{align*}\left(\sqrt{3}, -1\right)\end{align*}.

### Examples

#### Example 1

Earlier, you were asked to find the Cartesian coordinates for a point whose Polar coordinates are \begin{align*}(2, 330^\circ)\end{align*}.

First draw a diagram. From the diagram we can see that the reference angle is \begin{align*}30^\circ\end{align*}. Now we can use right triangle trigonometry to find \begin{align*}x\end{align*} and \begin{align*}y\end{align*}. In this particular case, we can also use special right triangle ratios or the unit circle.

\begin{align*}\cos 30^\circ &=\frac{x}{2} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \sin 30^\circ =\frac{y}{2} \\ x &=2 \cos 30^\circ=2 \left(\frac{\sqrt{3}}{2} \right)=\sqrt{3} \qquad and \qquad \ \ \qquad y =2 \cos 30^\circ =2 \left(\frac{1}{2} \right)=1\end{align*}

Since the point is in the fourth quadrant, the \begin{align*}y\end{align*} value should be negative giving the Cartesian coordinates \begin{align*}\left(\sqrt{3}, -1\right)\end{align*}.

#### Example 2

Use your calculator to find the Cartesian coordinates equivalent to the Polar coordinates \begin{align*}(11, 157^\circ)\end{align*}.

\begin{align*}(11 \cos 157^\circ, 11 \sin 157^\circ) \approx (-10.13, 4.30)\end{align*}

#### Example 3

Find the exact value of the Cartesian coordinates equivalent to the Polar coordinates \begin{align*}(8, 45^\circ)\end{align*}.

\begin{align*}(8 \cos 45^\circ, 8 \sin 45^\circ)=\left(8 \left(\frac{\sqrt{2}}{2} \right), 8 \left(\frac{\sqrt{2}}{2} \right) \right)=(4 \sqrt{2}, 4 \sqrt{2})\end{align*}

#### Example 4

Find the exact value of the Cartesian coordinates equivalent to the Polar coordinates \begin{align*}\left(5, - \frac{\pi}{2} \right)\end{align*}.

\begin{align*}\left(5 \cos \left(-\frac{\pi}{2} \right), 5 \sin \left(-\frac{\pi}{2} \right) \right)=(5(0), 5(-1))=(0, -5)\end{align*}

### Review

Use your calculator to find the Cartesian coordinates equivalent to the following Polar coordinates. Give your answers rounded to the nearest hundredth.

1. \begin{align*}(13, 38^\circ)\end{align*}
2. \begin{align*}(25, -230^\circ)\end{align*}
3. \begin{align*}(17, 345^\circ)\end{align*}
4. \begin{align*}(2, 140^\circ)\end{align*}
5. \begin{align*}\left(7, \frac{2 \pi}{5} \right)\end{align*}
6. \begin{align*}(9, 2.98)\end{align*}
7. \begin{align*}(3, -5.87)\end{align*}
8. \begin{align*}\left(10, \frac{13 \pi}{7} \right)\end{align*}

Find the exact value Cartesian coordinates equivalent to the following Polar coordinates.

1. \begin{align*}\left(5, \frac{\pi}{3} \right)\end{align*}
2. \begin{align*}\left(6, -\frac{\pi}{4} \right)\end{align*}
3. \begin{align*}\left(12, \frac{5 \pi}{6} \right)\end{align*}
4. \begin{align*}(7, \pi)\end{align*}
5. \begin{align*}(11, 2 \pi)\end{align*}
6. \begin{align*}\left(14, \frac{4 \pi}{3} \right)\end{align*}
7. \begin{align*}\left(27, \frac{3 \pi}{4} \right)\end{align*}
8. \begin{align*}\left(40, -\frac{5 \pi}{6} \right)\end{align*}

To see the Review answers, open this PDF file and look for section 13.11.

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