Trigonometry Examples

(4, - 4)

$(4, - 4)$

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{{(4)}^{2} + {(- 4)}^{2}}

$r = \sqrt{{(4)}^{2} + {(- 4)}^{2}}$

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.

Tap for more steps...

Step 3.1

Raise

4

$4$ to the power of

2

$2$ .

r = \sqrt{16 + {(- 4)}^{2}}

$r = \sqrt{16 + {(- 4)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Raise

- 4

$- 4$ to the power of

2

$2$ .

r = \sqrt{16 + 16}

$r = \sqrt{16 + 16}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Add

16

$16$ and

16

$16$ .

r = \sqrt{32}

$r = \sqrt{32}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Rewrite

32

$32$ as

4^{2} \cdot 2

$4^{2} \cdot 2$ .

Tap for more steps...

Step 3.4.1

Factor

16

$16$ out of

32

$32$ .

r = \sqrt{16 (2)}

$r = \sqrt{16 (2)}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4.2

Rewrite

16

$16$ as

4^{2}

$4^{2}$ .

r = \sqrt{4^{2} \cdot 2}

$r = \sqrt{4^{2} \cdot 2}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

r = \sqrt{4^{2} \cdot 2}

$r = \sqrt{4^{2} \cdot 2}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

Pull terms out from under the radical.

r = 4 \sqrt{2}

$r = 4 \sqrt{2}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

r = 4 \sqrt{2}

$r = 4 \sqrt{2}$

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 = 4 \sqrt{2}

$r = 4 \sqrt{2}$

θ = t a n^{- 1} (\frac{- 4}{4})

$θ = t a n^{- 1} (\frac{- 4}{4})$

Step 5

The inverse tangent of

- 1

$- 1$ is

θ = 315 °

$θ = 315 °$ .

r = 4 \sqrt{2}

$r = 4 \sqrt{2}$

θ = 315 °

$θ = 315 °$

Step 6

This is the result of the conversion to polar coordinates in

(r, θ)

$(r, θ)$ form.

(4 \sqrt{2}, 315 °)

$(4 \sqrt{2}, 315 °)$

⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

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