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Trigonometry Examples

(- 9, - 7)

$(- 9, - 7)$

Step 1

Convert from rectangular coordinates

(x, y)

$(x, y)$ to polar coordinates

(r, θ)

$(r, θ)$ using the conversion formulas.

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

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

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 2

Replace

x

$x$ and

y

$y$ with the actual values.

r = \sqrt{{(- 9)}^{2} + {(- 7)}^{2}}

$r = \sqrt{{(- 9)}^{2} + {(- 7)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Raise

- 9

$- 9$ to the power of

2

$2$ .

r = \sqrt{81 + {(- 7)}^{2}}

$r = \sqrt{81 + {(- 7)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Raise

- 7

$- 7$ to the power of

2

$2$ .

r = \sqrt{81 + 49}

$r = \sqrt{81 + 49}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add

81

$81$ and

49

$49$ .

r = \sqrt{130}

$r = \sqrt{130}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

r = \sqrt{130}

$r = \sqrt{130}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

Replace

x

$x$ and

y

$y$ with the actual values.

r = \sqrt{130}

$r = \sqrt{130}$

θ = t a n^{- 1} (\frac{- 7}{- 9})

$θ = t a n^{- 1} (\frac{- 7}{- 9})$

Step 5

The inverse tangent of

\frac{7}{9}

$\frac{7}{9}$ is

θ = 217.87498365 °

$θ = 217.87498365 °$ .

r = \sqrt{130}

$r = \sqrt{130}$

θ = 217.87498365 °

$θ = 217.87498365 °$

Step 6

This is the result of the conversion to polar coordinates in

(r, θ)

$(r, θ)$ form.

(\sqrt{130}, 217.87498365 °)

$(\sqrt{130}, 217.87498365 °)$

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

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