Precalculus Examples

$(1, \frac{11 π}{6})$

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 $θ = \frac{11 π}{6}$ into the formulas.

$x = (1) \cos (\frac{11 π}{6})$

$y = (1) \sin (\frac{11 π}{6})$

Step 3

Multiply $\cos (\frac{11 π}{6})$ by $1$ .

$x = \cos (\frac{11 π}{6})$

$y = (1) \sin (\frac{11 π}{6})$

Step 4

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

$x = \cos (\frac{π}{6})$

$y = (1) \sin (\frac{11 π}{6})$

Step 5

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

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

$y = (1) \sin (\frac{11 π}{6})$

Step 6

Multiply $\sin (\frac{11 π}{6})$ by $1$ .

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

$y = \sin (\frac{11 π}{6})$

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.

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

$y = - \sin (\frac{π}{6})$

Step 8

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

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

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

Step 9

The rectangular representation of the polar point $(1, \frac{11 π}{6})$ is $(\frac{\sqrt{3}}{2}, - \frac{1}{2})$ .

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