Trigonometry Examples

$(\sqrt{3}, - 1)$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

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

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Step 3.1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.3

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.4

Cancel the common factor of $2$ .

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Step 3.1.4.1

Cancel the common factor.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.4.2

Rewrite the expression.

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.1.5

Evaluate the exponent.

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

Simplify the expression.

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Step 3.2.1

Raise $- 1$ to the power of $2$ .

$r = \sqrt{3 + 1}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

Add $3$ and $1$ .

$r = \sqrt{4}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.3

Rewrite $4$ as $2^{2}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

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

$r = 2$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 2$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 2$

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

Step 5

The inverse tangent of $- \frac{\sqrt{3}}{3}$ is $θ = 330 °$ .

$r = 2$

$θ = 330 °$

Step 6

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

$(2, 330 °)$