Trigonometry Examples

$(- 4, - \frac{π}{2})$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

Raise $- 4$ to the power of $2$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2

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

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Step 3.2.1

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

$r = \sqrt{16 + {(- 1)}^{2} {(\frac{π}{2})}^{2}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

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

$r = \sqrt{16 + {(- 1)}^{2} (\frac{π^{2}}{2^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{16 + {(- 1)}^{2} (\frac{π^{2}}{2^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

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

$r = \sqrt{16 + 1 (\frac{π^{2}}{2^{2}})}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

Multiply $\frac{π^{2}}{2^{2}}$ by $1$ .

$r = \sqrt{16 + \frac{π^{2}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

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

$r = \sqrt{16 + \frac{π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

To write $16$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$r = \sqrt{16 \cdot \frac{4}{4} + \frac{π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Combine $16$ and $\frac{4}{4}$ .

$r = \sqrt{\frac{16 \cdot 4}{4} + \frac{π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Simplify the expression.

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Step 3.8.1

Combine the numerators over the common denominator.

$r = \sqrt{\frac{16 \cdot 4 + π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8.2

Multiply $16$ by $4$ .

$r = \sqrt{\frac{64 + π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{64 + π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

Rewrite $\sqrt{\frac{64 + π^{2}}{4}}$ as $\frac{\sqrt{64 + π^{2}}}{\sqrt{4}}$ .

$r = \frac{\sqrt{64 + π^{2}}}{\sqrt{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10

Simplify the denominator.

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Step 3.10.1

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

$r = \frac{\sqrt{64 + π^{2}}}{\sqrt{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.10.2

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

$r = \frac{\sqrt{64 + π^{2}}}{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{64 + π^{2}}}{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{64 + π^{2}}}{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{64 + π^{2}}}{2}$

$θ = t a n^{- 1} (\frac{- \frac{π}{2}}{- 4})$

Step 5

The inverse tangent of $\frac{π}{8}$ is $θ = 201.4398905 °$ .

$r = \frac{\sqrt{64 + π^{2}}}{2}$

$θ = 201.4398905 °$

Step 6

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

$(\frac{\sqrt{64 + π^{2}}}{2}, 201.4398905 °)$