Trigonometry Examples

$(2, \frac{3 π}{2})$

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.

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3

Find the magnitude of the polar coordinate.

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Step 3.1

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

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

$θ = t a n^{- 1} (\frac{y}{x})$

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 $\frac{3 π}{2}$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.2.2

Apply the product rule to $3 π$ .

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

$θ = t a n^{- 1} (\frac{y}{x})$

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

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.3

Raise $3$ to the power of $2$ .

$r = \sqrt{4 + \frac{9 π^{2}}{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.4

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

$r = \sqrt{4 + \frac{9 π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.5

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

$r = \sqrt{4 \cdot \frac{4}{4} + \frac{9 π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.6

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

$r = \sqrt{\frac{4 \cdot 4}{4} + \frac{9 π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7

Simplify the expression.

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Step 3.7.1

Combine the numerators over the common denominator.

$r = \sqrt{\frac{4 \cdot 4 + 9 π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.7.2

Multiply $4$ by $4$ .

$r = \sqrt{\frac{16 + 9 π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \sqrt{\frac{16 + 9 π^{2}}{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.8

Rewrite $\sqrt{\frac{16 + 9 π^{2}}{4}}$ as $\frac{\sqrt{16 + 9 π^{2}}}{\sqrt{4}}$ .

$r = \frac{\sqrt{16 + 9 π^{2}}}{\sqrt{4}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9

Simplify the denominator.

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Step 3.9.1

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

$r = \frac{\sqrt{16 + 9 π^{2}}}{\sqrt{2^{2}}}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 3.9.2

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

$r = \frac{\sqrt{16 + 9 π^{2}}}{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{16 + 9 π^{2}}}{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

$r = \frac{\sqrt{16 + 9 π^{2}}}{2}$

$θ = t a n^{- 1} (\frac{y}{x})$

Step 4

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

$r = \frac{\sqrt{16 + 9 π^{2}}}{2}$

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

Step 5

The inverse tangent of $\frac{3 π}{4}$ is $θ = 67.00299232 °$ .

$r = \frac{\sqrt{16 + 9 π^{2}}}{2}$

$θ = 67.00299232 °$

Step 6

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

$(\frac{\sqrt{16 + 9 π^{2}}}{2}, 67.00299232 °)$