Trigonometry Examples

$(- 9, - 9)$

Step 1

Convert from rectangular coordinates $(x, y)$ to polar coordinates $(r, θ)$ using the conversion formulas.

$r = \sqrt{x^{2} + y^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 2

Replace $x$ and $y$ with the actual values.

$r = \sqrt{{(- 9)}^{2} + {(- 9)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Raise $- 9$ to the power of $2$ .

$r = \sqrt{81 + {(- 9)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Raise $- 9$ to the power of $2$ .

$r = \sqrt{81 + 81}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add $81$ and $81$ .

$r = \sqrt{162}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Rewrite $162$ as $9^{2} \cdot 2$ .

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Step 3.4.1

Factor $81$ out of $162$ .

$r = \sqrt{81 (2)}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

Rewrite $81$ as $9^{2}$ .

$r = \sqrt{9^{2} \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9^{2} \cdot 2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Pull terms out from under the radical.

$r = 9 \sqrt{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 9 \sqrt{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

Replace $x$ and $y$ with the actual values.

$r = 9 \sqrt{2}$

$θ = t a n^{- 1} (\frac{- 9}{- 9})$

Step 5

The inverse tangent of $1$ is $θ = 225 °$ .

$r = 9 \sqrt{2}$

$θ = 225 °$

Step 6

This is the result of the conversion to polar coordinates in $(r, θ)$ form.

$(9 \sqrt{2}, 225 °)$