Precalculus Examples

$(- 3, \frac{7 π}{6})$

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 $r = - 3$ and $θ = \frac{7 π}{6}$ into the formulas.

$x = (- 3) \cos (\frac{7 π}{6})$

$y = (- 3) \sin (\frac{7 π}{6})$

Step 3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the third quadrant.

$x = - 3 (- \cos (\frac{π}{6}))$

$y = (- 3) \sin (\frac{7 π}{6})$

Step 4

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

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

$y = (- 3) \sin (\frac{7 π}{6})$

Step 5

Multiply $- 3 (- \frac{\sqrt{3}}{2})$ .

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Step 5.1

Multiply $- 1$ by $- 3$ .

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

$y = (- 3) \sin (\frac{7 π}{6})$

Step 5.2

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

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

$y = (- 3) \sin (\frac{7 π}{6})$

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

$y = (- 3) \sin (\frac{7 π}{6})$

Step 6

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the third quadrant.

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

$y = - 3 (- \sin (\frac{π}{6}))$

Step 7

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

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

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

Step 8

Multiply $- 3 (- \frac{1}{2})$ .

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Step 8.1

Multiply $- 1$ by $- 3$ .

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

$y = 3 (\frac{1}{2})$

Step 8.2

Combine $3$ and $\frac{1}{2}$ .

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

$y = \frac{3}{2}$

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

$y = \frac{3}{2}$

Step 9

The rectangular representation of the polar point $(- 3, \frac{7 π}{6})$ is $(\frac{3 \sqrt{3}}{2}, \frac{3}{2})$ .

$(\frac{3 \sqrt{3}}{2}, \frac{3}{2})$