Trigonometry Examples

$(\sqrt{3}, π)$

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

$x = (\sqrt{3}) \cos (π)$

$y = (\sqrt{3}) \sin (π)$

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.

$x = \sqrt{3} (- \cos (0))$

$y = (\sqrt{3}) \sin (π)$

Step 4

The exact value of $\cos (0)$ is $1$ .

$x = \sqrt{3} (- 1 \cdot 1)$

$y = (\sqrt{3}) \sin (π)$

Step 5

Multiply $- 1$ by $1$ .

$x = \sqrt{3} \cdot - 1$

$y = (\sqrt{3}) \sin (π)$

Step 6

Move $- 1$ to the left of $\sqrt{3}$ .

$x = - 1 \cdot \sqrt{3}$

$y = (\sqrt{3}) \sin (π)$

Step 7

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

$x = - \sqrt{3}$

$y = (\sqrt{3}) \sin (π)$

Step 8

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

$x = - \sqrt{3}$

$y = \sqrt{3} \sin (0)$

Step 9

The exact value of $\sin (0)$ is $0$ .

$x = - \sqrt{3}$

$y = \sqrt{3} \cdot 0$

Step 10

Multiply $\sqrt{3}$ by $0$ .

$x = - \sqrt{3}$

$y = 0$

Step 11

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

$(- \sqrt{3}, 0)$