Precalculus Examples

$(5 \sqrt{2}, - \frac{11 π}{6})$

Step 1

Use the conversion formulas to convert from polar coordinates to rectangular coordinates.

$x = r c o s θ$

$y = r s i n θ$

Step 2

Substitute in the known values of $r = 5 \sqrt{2}$ and $θ = - \frac{11 π}{6}$ into the formulas.

$x = (5 \sqrt{2}) \cos (- \frac{11 π}{6})$

$y = (5 \sqrt{2}) \sin (- \frac{11 π}{6})$

Step 3

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$x = 5 \sqrt{2} \cos (\frac{π}{6})$

$y = (5 \sqrt{2}) \sin (- \frac{11 π}{6})$

Step 4

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$x = 5 \sqrt{2} (\frac{\sqrt{3}}{2})$

$y = (5 \sqrt{2}) \sin (- \frac{11 π}{6})$

Step 5

Multiply $5 \sqrt{2} \frac{\sqrt{3}}{2}$ .

Tap for more steps...

Step 5.1

Combine $\frac{\sqrt{3}}{2}$ and $5$ .

$x = \frac{\sqrt{3} \cdot 5}{2} \cdot \sqrt{2}$

$y = (5 \sqrt{2}) \sin (- \frac{11 π}{6})$

Step 5.2

Combine $\frac{\sqrt{3} \cdot 5}{2}$ and $\sqrt{2}$ .

$x = \frac{\sqrt{3} \cdot (5 \sqrt{2})}{2}$

$y = (5 \sqrt{2}) \sin (- \frac{11 π}{6})$

Step 5.3

Combine using the product rule for radicals.

$x = \frac{5 \sqrt{3 \cdot 2}}{2}$

$y = (5 \sqrt{2}) \sin (- \frac{11 π}{6})$

Step 5.4

Multiply $3$ by $2$ .

$x = \frac{5 \sqrt{6}}{2}$

$y = (5 \sqrt{2}) \sin (- \frac{11 π}{6})$

$x = \frac{5 \sqrt{6}}{2}$

$y = (5 \sqrt{2}) \sin (- \frac{11 π}{6})$

Step 6

Add full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$x = \frac{5 \sqrt{6}}{2}$

$y = 5 \sqrt{2} \sin (\frac{π}{6})$

Step 7

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$x = \frac{5 \sqrt{6}}{2}$

$y = 5 \sqrt{2} (\frac{1}{2})$

Step 8

Multiply $5 \sqrt{2} \frac{1}{2}$ .

Tap for more steps...

Step 8.1

Combine $\frac{1}{2}$ and $5$ .

$x = \frac{5 \sqrt{6}}{2}$

$y = \frac{5}{2} \cdot \sqrt{2}$

Step 8.2

Combine $\frac{5}{2}$ and $\sqrt{2}$ .

$x = \frac{5 \sqrt{6}}{2}$

$y = \frac{5 \sqrt{2}}{2}$

$x = \frac{5 \sqrt{6}}{2}$

$y = \frac{5 \sqrt{2}}{2}$

Step 9

The rectangular representation of the polar point $(5 \sqrt{2}, - \frac{11 π}{6})$ is $(\frac{5 \sqrt{6}}{2}, \frac{5 \sqrt{2}}{2})$ .

$(\frac{5 \sqrt{6}}{2}, \frac{5 \sqrt{2}}{2})$