Trigonometry Examples

$(- 5, - \frac{π}{3})$

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 $r = - 5$ and $θ = - \frac{π}{3}$ into the formulas.

$x = (- 5) \cos (- \frac{π}{3})$

$y = (- 5) \sin (- \frac{π}{3})$

Step 3

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

$x = - 5 \cos (\frac{5 π}{3})$

$y = (- 5) \sin (- \frac{π}{3})$

Step 4

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

$x = - 5 \cos (\frac{π}{3})$

$y = (- 5) \sin (- \frac{π}{3})$

Step 5

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

$x = - 5 (\frac{1}{2})$

$y = (- 5) \sin (- \frac{π}{3})$

Step 6

Combine $- 5$ and $\frac{1}{2}$ .

$x = \frac{- 5}{2}$

$y = (- 5) \sin (- \frac{π}{3})$

Step 7

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

$y = (- 5) \sin (- \frac{π}{3})$

Step 8

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

$x = - \frac{5}{2}$

$y = - 5 \sin (\frac{5 π}{3})$

Step 9

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.

$x = - \frac{5}{2}$

$y = - 5 (- \sin (\frac{π}{3}))$

Step 10

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

$x = - \frac{5}{2}$

$y = - 5 (- \frac{\sqrt{3}}{2})$

Step 11

Multiply $- 5 (- \frac{\sqrt{3}}{2})$ .

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Step 11.1

Multiply $- 1$ by $- 5$ .

$x = - \frac{5}{2}$

$y = 5 (\frac{\sqrt{3}}{2})$

Step 11.2

Combine $5$ and $\frac{\sqrt{3}}{2}$ .

$x = - \frac{5}{2}$

$y = \frac{5 \sqrt{3}}{2}$

$x = - \frac{5}{2}$

$y = \frac{5 \sqrt{3}}{2}$

Step 12

The rectangular representation of the polar point $(- 5, - \frac{π}{3})$ is $(- \frac{5}{2}, \frac{5 \sqrt{3}}{2})$ .

$(- \frac{5}{2}, \frac{5 \sqrt{3}}{2})$

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