Precalculus Examples

$(- 7, \frac{2 π}{3})$

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 $r = - 7$ and $θ = \frac{2 π}{3}$ into the formulas.

$x = (- 7) \cos (\frac{2 π}{3})$

$y = (- 7) \sin (\frac{2 π}{3})$

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 second quadrant.

$x = - 7 (- \cos (\frac{π}{3}))$

$y = (- 7) \sin (\frac{2 π}{3})$

Step 4

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$x = - 7 (- \frac{1}{2})$

$y = (- 7) \sin (\frac{2 π}{3})$

Step 5

Multiply $- 7 (- \frac{1}{2})$ .

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Step 5.1

Multiply $- 1$ by $- 7$ .

$x = 7 (\frac{1}{2})$

$y = (- 7) \sin (\frac{2 π}{3})$

Step 5.2

Combine $7$ and $\frac{1}{2}$ .

$x = \frac{7}{2}$

$y = (- 7) \sin (\frac{2 π}{3})$

$x = \frac{7}{2}$

$y = (- 7) \sin (\frac{2 π}{3})$

Step 6

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$x = \frac{7}{2}$

$y = - 7 \sin (\frac{π}{3})$

Step 7

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

$x = \frac{7}{2}$

$y = - 7 \frac{\sqrt{3}}{2}$

Step 8

Combine $- 7$ and $\frac{\sqrt{3}}{2}$ .

$x = \frac{7}{2}$

$y = \frac{- 7 \sqrt{3}}{2}$

Step 9

Move the negative in front of the fraction.

$x = \frac{7}{2}$

$y = - \frac{7 \sqrt{3}}{2}$

Step 10

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

$(\frac{7}{2}, - \frac{7 \sqrt{3}}{2})$