Trigonometry Examples

$(\frac{\sqrt{3}}{2}, \frac{1}{2})$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

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Step 3.2.1

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.3

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.4

Cancel the common factor of $2$ .

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Step 3.2.4.1

Cancel the common factor.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.4.2

Rewrite the expression.

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.5

Evaluate the exponent.

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Simplify the expression.

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Step 3.3.1

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.2

Apply the product rule to $\frac{1}{2}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.3

One to any power is one.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.4

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

$r = \sqrt{\frac{3}{4} + \frac{1}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.5

Combine the numerators over the common denominator.

$r = \sqrt{\frac{3 + 1}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.6

Add $3$ and $1$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.7

Divide $4$ by $4$ .

$r = \sqrt{1}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3.8

Any root of $1$ is $1$ .

$r = 1$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 1$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = 1$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = 1$

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

Step 5

The inverse tangent of $\frac{\sqrt{3}}{3}$ is $θ = 30 °$ .

$r = 1$

$θ = 30 °$

Step 6

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

$(1, 30 °)$