Algebra Examples

$(\sqrt{2}, \frac{3 π}{2})$

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

$x = (\sqrt{2}) \cos (\frac{3 π}{2})$

$y = (\sqrt{2}) \sin (\frac{3 π}{2})$

Step 3

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

$x = \sqrt{2} \cos (\frac{π}{2})$

$y = (\sqrt{2}) \sin (\frac{3 π}{2})$

Step 4

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$x = \sqrt{2} \cdot 0$

$y = (\sqrt{2}) \sin (\frac{3 π}{2})$

Step 5

Multiply $\sqrt{2}$ by $0$ .

$x = 0$

$y = (\sqrt{2}) \sin (\frac{3 π}{2})$

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 fourth quadrant.

$x = 0$

$y = \sqrt{2} (- \sin (\frac{π}{2}))$

Step 7

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

$x = 0$

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

Step 8

Multiply $- 1$ by $1$ .

$x = 0$

$y = \sqrt{2} \cdot - 1$

Step 9

Move $- 1$ to the left of $\sqrt{2}$ .

$x = 0$

$y = - 1 \cdot \sqrt{2}$

Step 10

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

$x = 0$

$y = - \sqrt{2}$

Step 11

The rectangular representation of the polar point $(\sqrt{2}, \frac{3 π}{2})$ is $(0, - \sqrt{2})$ .

$(0, - \sqrt{2})$