Trigonometry Examples

$(- 9, \frac{4 π}{3})$

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 $r = - 9$ and $θ = \frac{4 π}{3}$ into the formulas.

$x = (- 9) \cos (\frac{4 π}{3})$

$y = (- 9) \sin (\frac{4 π}{3})$

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 third quadrant.

$x = - 9 (- \cos (\frac{π}{3}))$

$y = (- 9) \sin (\frac{4 π}{3})$

Step 4

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

$x = - 9 (- \frac{1}{2})$

$y = (- 9) \sin (\frac{4 π}{3})$

Step 5

Multiply $- 9 (- \frac{1}{2})$ .

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Step 5.1

Multiply $- 1$ by $- 9$ .

$x = 9 (\frac{1}{2})$

$y = (- 9) \sin (\frac{4 π}{3})$

Step 5.2

Combine $9$ and $\frac{1}{2}$ .

$x = \frac{9}{2}$

$y = (- 9) \sin (\frac{4 π}{3})$

$x = \frac{9}{2}$

$y = (- 9) \sin (\frac{4 π}{3})$

Step 6

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 third quadrant.

$x = \frac{9}{2}$

$y = - 9 (- \sin (\frac{π}{3}))$

Step 7

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

$x = \frac{9}{2}$

$y = - 9 (- \frac{\sqrt{3}}{2})$

Step 8

Multiply $- 9 (- \frac{\sqrt{3}}{2})$ .

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Step 8.1

Multiply $- 1$ by $- 9$ .

$x = \frac{9}{2}$

$y = 9 (\frac{\sqrt{3}}{2})$

Step 8.2

Combine $9$ and $\frac{\sqrt{3}}{2}$ .

$x = \frac{9}{2}$

$y = \frac{9 \sqrt{3}}{2}$

$x = \frac{9}{2}$

$y = \frac{9 \sqrt{3}}{2}$

Step 9

The rectangular representation of the polar point $(- 9, \frac{4 π}{3})$ is $(\frac{9}{2}, \frac{9 \sqrt{3}}{2})$ .

$(\frac{9}{2}, \frac{9 \sqrt{3}}{2})$