Trigonometry Examples

$(\frac{5}{2}, \frac{5 \sqrt{3}}{2})$

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.

$r = \sqrt{{(\frac{5}{2})}^{2} + {(\frac{5 \sqrt{3}}{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Apply the product rule to $\frac{5}{2}$ .

$r = \sqrt{\frac{5^{2}}{2^{2}} + {(\frac{5 \sqrt{3}}{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Raise $5$ to the power of $2$ .

$r = \sqrt{\frac{25}{2^{2}} + {(\frac{5 \sqrt{3}}{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

$r = \sqrt{\frac{25}{4} + {(\frac{5 \sqrt{3}}{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Apply the product rule to $\frac{5 \sqrt{3}}{2}$ .

$r = \sqrt{\frac{25}{4} + \frac{{(5 \sqrt{3})}^{2}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Apply the product rule to $5 \sqrt{3}$ .

$r = \sqrt{\frac{25}{4} + \frac{5^{2} {\sqrt{3}}^{2}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{25}{4} + \frac{5^{2} {\sqrt{3}}^{2}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Simplify the numerator.

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Raise $5$ to the power of $2$ .

$r = \sqrt{\frac{25}{4} + \frac{25 {\sqrt{3}}^{2}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

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

$r = \sqrt{\frac{25}{4} + \frac{25 {(3^{\frac{1}{2}})}^{2}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

$r = \sqrt{\frac{25}{4} + \frac{25 \cdot 3^{\frac{1}{2} \cdot 2}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

$r = \sqrt{\frac{25}{4} + \frac{25 \cdot 3^{\frac{2}{2}}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Cancel the common factor of $2$ .

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Cancel the common factor.

$r = \sqrt{\frac{25}{4} + \frac{25 \cdot 3^{\frac{2}{2}}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite the expression.

$r = \sqrt{\frac{25}{4} + \frac{25 \cdot 3}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{25}{4} + \frac{25 \cdot 3}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Evaluate the exponent.

$r = \sqrt{\frac{25}{4} + \frac{25 \cdot 3}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{25}{4} + \frac{25 \cdot 3}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{25}{4} + \frac{25 \cdot 3}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Simplify the expression.

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

$r = \sqrt{\frac{25}{4} + \frac{25 \cdot 3}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Multiply $25$ by $3$ .

$r = \sqrt{\frac{25}{4} + \frac{75}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Combine the numerators over the common denominator.

$r = \sqrt{\frac{25 + 75}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Add $25$ and $75$ .

$r = \sqrt{\frac{100}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Divide $100$ by $4$ .

$r = \sqrt{25}$

$θ = t a n^{- 1} (\frac{y}{x})$

Rewrite $25$ as $5^{2}$ .

$r = \sqrt{5^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{5^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

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

$r = 5$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 5$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 5$

$θ = t a n^{- 1} (\frac{\frac{5 \sqrt{3}}{2}}{\frac{5}{2}})$

Step 5

The inverse tangent of $\sqrt{3}$ is $θ = 60 °$ .

$r = 5$

$θ = 60 °$

Step 6

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

$(5, 60 °)$