Trigonometry Examples

(- 3 \sqrt{3}, 3)

$(- 3 \sqrt{3}, 3)$

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{{(- 3 \sqrt{3})}^{2} + {(3)}^{2}}

$r = \sqrt{{(- 3 \sqrt{3})}^{2} + {(3)}^{2}}$

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

Simplify the expression.

Tap for more steps...

Step 3.1.1

Apply the product rule to

- 3 \sqrt{3}

$- 3 \sqrt{3}$ .

r = \sqrt{{(- 3)}^{2} {\sqrt{3}}^{2} + {(3)}^{2}}

$r = \sqrt{{(- 3)}^{2} {\sqrt{3}}^{2} + {(3)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.2

Raise

- 3

$- 3$ to the power of

2

$2$ .

r = \sqrt{9 {\sqrt{3}}^{2} + {(3)}^{2}}

$r = \sqrt{9 {\sqrt{3}}^{2} + {(3)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

r = \sqrt{9 {\sqrt{3}}^{2} + {(3)}^{2}}

$r = \sqrt{9 {\sqrt{3}}^{2} + {(3)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Rewrite

{\sqrt{3}}^{2}

${\sqrt{3}}^{2}$ as

3

$3$ .

Tap for more steps...

Step 3.2.1

Use

\sqrt[n]{a^{x}} = a^{\frac{x}{n}}

$\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite

\sqrt{3}

$\sqrt{3}$ as

3^{\frac{1}{2}}

$3^{\frac{1}{2}}$ .

r = \sqrt{9 {(3^{\frac{1}{2}})}^{2} + {(3)}^{2}}

$r = \sqrt{9 {(3^{\frac{1}{2}})}^{2} + {(3)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

r = \sqrt{9 \cdot 3^{\frac{1}{2} \cdot 2} + {(3)}^{2}}

$r = \sqrt{9 \cdot 3^{\frac{1}{2} \cdot 2} + {(3)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.3

Combine

\frac{1}{2}

$\frac{1}{2}$ and

2

$2$ .

r = \sqrt{9 \cdot 3^{\frac{2}{2}} + {(3)}^{2}}

$r = \sqrt{9 \cdot 3^{\frac{2}{2}} + {(3)}^{2}}$

θ = t a n^{- 1} (\frac{y}{x})

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.4

Cancel the common factor of

2

$2$ .

Tap for more steps...

Step 3.2.4.1

Cancel the common factor.

$r = \sqrt{9 \cdot 3^{\frac{2}{2}} + {(3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.4.2

Rewrite the expression.

$r = \sqrt{9 \cdot 3 + {(3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9 \cdot 3 + {(3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.5

Evaluate the exponent.

$r = \sqrt{9 \cdot 3 + {(3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{9 \cdot 3 + {(3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Simplify the expression.

Tap for more steps...

Step 3.3.1

Multiply

$9$ by

$3$ .

$r = \sqrt{27 + {(3)}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.2

Raise

$3$ to the power of

$2$ .

$r = \sqrt{27 + 9}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.3

Add

$27$ and

$9$ .

$r = \sqrt{36}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.4

Rewrite

$36$ as

$6^{2}$ .

$r = \sqrt{6^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.5

Pull terms out from under the radical, assuming positive real numbers.

$r = 6$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 6$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 6$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

Replace

$x$ and

$y$ with the actual values.

$r = 6$

$θ = t a n^{- 1} (\frac{3}{- 3 \sqrt{3}})$

Step 5

The inverse tangent of

$- \frac{\sqrt{3}}{3}$ is

$θ = 150 °$ .

$r = 6$

$θ = 150 °$

Step 6

This is the result of the conversion to polar coordinates in

$(r, θ)$ form.

$(6, 150 °)$