Precalculus Examples

$(- 3, \frac{5 π}{3})$

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 $r = - 3$ and $θ = \frac{5 π}{3}$ into the formulas.

$x = (- 3) \cos (\frac{5 π}{3})$

$y = (- 3) \sin (\frac{5 π}{3})$

Step 3

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

$x = - 3 \cos (\frac{π}{3})$

$y = (- 3) \sin (\frac{5 π}{3})$

Step 4

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

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

$y = (- 3) \sin (\frac{5 π}{3})$

Step 5

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

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

$y = (- 3) \sin (\frac{5 π}{3})$

Step 6

Move the negative in front of the fraction.

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

$y = (- 3) \sin (\frac{5 π}{3})$

Step 7

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 fourth quadrant.

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

$y = - 3 (- \sin (\frac{π}{3}))$

Step 8

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

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

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

Step 9

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

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Step 9.1

Multiply $- 1$ by $- 3$ .

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

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

Step 9.2

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

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

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

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

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

Step 10

The rectangular representation of the polar point $(- 3, \frac{5 π}{3})$ is $(- \frac{3}{2}, \frac{3 \sqrt{3}}{2})$ .

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

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$