Trigonometry Examples

$(- 1, - \frac{5 π}{3})$

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 $r = - 1$ and $θ = - \frac{5 π}{3}$ into the formulas.

$x = (- 1) \cos (- \frac{5 π}{3})$

$y = (- 1) \sin (- \frac{5 π}{3})$

Step 3

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

$x = - 1 \cos (\frac{π}{3})$

$y = (- 1) \sin (- \frac{5 π}{3})$

Step 4

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

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

$y = (- 1) \sin (- \frac{5 π}{3})$

Step 5

Rewrite $- 1 (\frac{1}{2})$ as $- (\frac{1}{2})$ .

$x = - \frac{1}{2}$

$y = (- 1) \sin (- \frac{5 π}{3})$

Step 6

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

$x = - \frac{1}{2}$

$y = - 1 \sin (\frac{π}{3})$

Step 7

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

$x = - \frac{1}{2}$

$y = - 1 \frac{\sqrt{3}}{2}$

Step 8

Rewrite $- 1 \frac{\sqrt{3}}{2}$ as $- \frac{\sqrt{3}}{2}$ .

$x = - \frac{1}{2}$

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

Step 9

The rectangular representation of the polar point $(- 1, - \frac{5 π}{3})$ is $(- \frac{1}{2}, - \frac{\sqrt{3}}{2})$ .

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