Precalculus Examples

$(- 3, \frac{- π}{2})$

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

$x = (- 3) \cos (\frac{- π}{2})$

$y = (- 3) \sin (\frac{- π}{2})$

Step 3

Move the negative in front of the fraction.

$x = - 3 \cos (- \frac{π}{2})$

$y = (- 3) \sin (\frac{- π}{2})$

Step 4

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

$x = - 3 \cos (\frac{3 π}{2})$

$y = (- 3) \sin (\frac{- π}{2})$

Step 5

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

$x = - 3 \cos (\frac{π}{2})$

$y = (- 3) \sin (\frac{- π}{2})$

Step 6

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

$x = - 3 \cdot 0$

$y = (- 3) \sin (\frac{- π}{2})$

Step 7

Multiply $- 3$ by $0$ .

$x = 0$

$y = (- 3) \sin (\frac{- π}{2})$

Step 8

Move the negative in front of the fraction.

$x = 0$

$y = - 3 \sin (- \frac{π}{2})$

Step 9

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

$x = 0$

$y = - 3 \sin (\frac{3 π}{2})$

Step 10

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.

$x = 0$

$y = - 3 (- \sin (\frac{π}{2}))$

Step 11

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

$x = 0$

$y = - 3 (- 1 \cdot 1)$

Step 12

Multiply $- 3 (- 1 \cdot 1)$ .

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Step 12.1

Multiply $- 1$ by $1$ .

$x = 0$

$y = - 3 \cdot - 1$

Step 12.2

Multiply $- 3$ by $- 1$ .

$x = 0$

$y = 3$

$x = 0$

$y = 3$

Step 13

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

$(0, 3)$

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