Trigonometry Examples

$(2, \frac{11 π}{3})$

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 $r = 2$ and $θ = \frac{11 π}{3}$ into the formulas.

$x = (2) \cos (\frac{11 π}{3})$

$y = (2) \sin (\frac{11 π}{3})$

Step 3

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

$x = 2 \cos (\frac{5 π}{3})$

$y = (2) \sin (\frac{11 π}{3})$

Step 4

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

$x = 2 \cos (\frac{π}{3})$

$y = (2) \sin (\frac{11 π}{3})$

Step 5

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

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

$y = (2) \sin (\frac{11 π}{3})$

Step 6

Cancel the common factor of $2$ .

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Step 6.1

Cancel the common factor.

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

$y = (2) \sin (\frac{11 π}{3})$

Step 6.2

Rewrite the expression.

$x = 1$

$y = (2) \sin (\frac{11 π}{3})$

$x = 1$

$y = (2) \sin (\frac{11 π}{3})$

Step 7

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

$x = 1$

$y = 2 \sin (\frac{5 π}{3})$

Step 8

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.

$x = 1$

$y = 2 (- \sin (\frac{π}{3}))$

Step 9

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

$x = 1$

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

Step 10

Cancel the common factor of $2$ .

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Step 10.1

Move the leading negative in $- \frac{\sqrt{3}}{2}$ into the numerator.

$x = 1$

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

Step 10.2

Cancel the common factor.

$x = 1$

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

Step 10.3

Rewrite the expression.

$x = 1$

$y = - \sqrt{3}$

$x = 1$

$y = - \sqrt{3}$

Step 11

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

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