Precalculus Examples

$(- 2, \frac{π}{6})$

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 $r = - 2$ and $θ = \frac{π}{6}$ into the formulas.

$x = (- 2) \cos (\frac{π}{6})$

$y = (- 2) \sin (\frac{π}{6})$

Step 3

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

$x = - 2 \frac{\sqrt{3}}{2}$

$y = (- 2) \sin (\frac{π}{6})$

Step 4

Cancel the common factor of $2$ .

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Step 4.1

Factor $2$ out of $- 2$ .

$x = 2 (- 1) (\frac{\sqrt{3}}{2})$

$y = (- 2) \sin (\frac{π}{6})$

Step 4.2

Cancel the common factor.

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

$y = (- 2) \sin (\frac{π}{6})$

Step 4.3

Rewrite the expression.

$x = - 1 \sqrt{3}$

$y = (- 2) \sin (\frac{π}{6})$

$x = - 1 \sqrt{3}$

$y = (- 2) \sin (\frac{π}{6})$

Step 5

Rewrite $- 1 \sqrt{3}$ as $- \sqrt{3}$ .

$x = - \sqrt{3}$

$y = (- 2) \sin (\frac{π}{6})$

Step 6

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

$x = - \sqrt{3}$

$y = - 2 (\frac{1}{2})$

Step 7

Cancel the common factor of $2$ .

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Step 7.1

Factor $2$ out of $- 2$ .

$x = - \sqrt{3}$

$y = 2 (- 1) (\frac{1}{2})$

Step 7.2

Cancel the common factor.

$x = - \sqrt{3}$

$y = 2 \cdot (- 1 (\frac{1}{2}))$

Step 7.3

Rewrite the expression.

$x = - \sqrt{3}$

$y = - 1$

$x = - \sqrt{3}$

$y = - 1$

Step 8

The rectangular representation of the polar point $(- 2, \frac{π}{6})$ is $(- \sqrt{3}, - 1)$ .

$(- \sqrt{3}, - 1)$