Precalculus Examples

$(4, \frac{π}{12})$

Step 1

Use the conversion formulas to convert from polar coordinates to rectangular coordinates.

$x = r c o s θ$

$y = r s i n θ$

Step 2

Substitute in the known values of $r = 4$ and $θ = \frac{π}{12}$ into the formulas.

$x = (4) \cos (\frac{π}{12})$

$y = (4) \sin (\frac{π}{12})$

Step 3

The exact value of $\cos (\frac{π}{12})$ is $\frac{\sqrt{6} + \sqrt{2}}{4}$ .

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Step 3.1

Split $\frac{π}{12}$ into two angles where the values of the six trigonometric functions are known.

$x = 4 \cos (\frac{π}{4} - \frac{π}{6})$

$y = (4) \sin (\frac{π}{12})$

Step 3.2

Apply the difference of angles identity $\cos (x - y) = \cos (x) \cos (y) + \sin (x) \sin (y)$ .

$x = 4 (\cos (\frac{π}{4}) \cos (\frac{π}{6}) + \sin (\frac{π}{4}) \sin (\frac{π}{6}))$

$y = (4) \sin (\frac{π}{12})$

Step 3.3

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$x = 4 (\frac{\sqrt{2}}{2} \cdot \cos (\frac{π}{6}) + \sin (\frac{π}{4}) \sin (\frac{π}{6}))$

$y = (4) \sin (\frac{π}{12})$

Step 3.4

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$x = 4 (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \sin (\frac{π}{4}) \sin (\frac{π}{6}))$

$y = (4) \sin (\frac{π}{12})$

Step 3.5

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$x = 4 (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \sin (\frac{π}{6}))$

$y = (4) \sin (\frac{π}{12})$

Step 3.6

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$x = 4 (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

$y = (4) \sin (\frac{π}{12})$

Step 3.7

Simplify $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$ .

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Step 3.7.1

Simplify each term.

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Step 3.7.1.1

Multiply $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}$ .

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Step 3.7.1.1.1

Multiply $\frac{\sqrt{2}}{2}$ by $\frac{\sqrt{3}}{2}$ .

$x = 4 (\frac{\sqrt{2} \sqrt{3}}{2 \cdot 2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

$y = (4) \sin (\frac{π}{12})$

Step 3.7.1.1.2

Combine using the product rule for radicals.

$x = 4 (\frac{\sqrt{2 \cdot 3}}{2 \cdot 2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

$y = (4) \sin (\frac{π}{12})$

Step 3.7.1.1.3

Multiply $2$ by $3$ .

$x = 4 (\frac{\sqrt{6}}{2 \cdot 2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

$y = (4) \sin (\frac{π}{12})$

Step 3.7.1.1.4

Multiply $2$ by $2$ .

$x = 4 (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

$y = (4) \sin (\frac{π}{12})$

$x = 4 (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

$y = (4) \sin (\frac{π}{12})$

Step 3.7.1.2

Multiply $\frac{\sqrt{2}}{2} \cdot \frac{1}{2}$ .

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Step 3.7.1.2.1

Multiply $\frac{\sqrt{2}}{2}$ by $\frac{1}{2}$ .

$x = 4 (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2 \cdot 2})$

$y = (4) \sin (\frac{π}{12})$

Step 3.7.1.2.2

Multiply $2$ by $2$ .

$x = 4 (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4})$

$y = (4) \sin (\frac{π}{12})$

$x = 4 (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4})$

$y = (4) \sin (\frac{π}{12})$

$x = 4 (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4})$

$y = (4) \sin (\frac{π}{12})$

Step 3.7.2

Combine the numerators over the common denominator.

$x = 4 (\frac{\sqrt{6} + \sqrt{2}}{4})$

$y = (4) \sin (\frac{π}{12})$

$x = 4 (\frac{\sqrt{6} + \sqrt{2}}{4})$

$y = (4) \sin (\frac{π}{12})$

$x = 4 (\frac{\sqrt{6} + \sqrt{2}}{4})$

$y = (4) \sin (\frac{π}{12})$

Step 4

Cancel the common factor of $4$ .

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Step 4.1

Cancel the common factor.

$x = 4 (\frac{\sqrt{6} + \sqrt{2}}{4})$

$y = (4) \sin (\frac{π}{12})$

Step 4.2

Rewrite the expression.

$x = \sqrt{6} + \sqrt{2}$

$y = (4) \sin (\frac{π}{12})$

$x = \sqrt{6} + \sqrt{2}$

$y = (4) \sin (\frac{π}{12})$

Step 5

The exact value of $\sin (\frac{π}{12})$ is $\frac{\sqrt{6} - \sqrt{2}}{4}$ .

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Step 5.1

Split $\frac{π}{12}$ into two angles where the values of the six trigonometric functions are known.

$x = \sqrt{6} + \sqrt{2}$

$y = 4 \sin (\frac{π}{4} - \frac{π}{6})$

Step 5.2

Apply the difference of angles identity.

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\sin (\frac{π}{4}) \cos (\frac{π}{6}) - \cos (\frac{π}{4}) \sin (\frac{π}{6}))$

Step 5.3

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{2}}{2} \cdot \cos (\frac{π}{6}) - \cos (\frac{π}{4}) \sin (\frac{π}{6}))$

Step 5.4

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \cos (\frac{π}{4}) \sin (\frac{π}{6}))$

Step 5.5

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \sin (\frac{π}{6}))$

Step 5.6

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

Step 5.7

Simplify $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$ .

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Step 5.7.1

Simplify each term.

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Step 5.7.1.1

Multiply $\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}$ .

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Step 5.7.1.1.1

Multiply $\frac{\sqrt{2}}{2}$ by $\frac{\sqrt{3}}{2}$ .

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{2} \sqrt{3}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

Step 5.7.1.1.2

Combine using the product rule for radicals.

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{2 \cdot 3}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

Step 5.7.1.1.3

Multiply $2$ by $3$ .

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

Step 5.7.1.1.4

Multiply $2$ by $2$ .

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2})$

Step 5.7.1.2

Multiply $- \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$ .

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Step 5.7.1.2.1

Multiply $\frac{1}{2}$ by $\frac{\sqrt{2}}{2}$ .

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{2 \cdot 2})$

Step 5.7.1.2.2

Multiply $2$ by $2$ .

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4})$

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4})$

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4})$

Step 5.7.2

Combine the numerators over the common denominator.

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6} - \sqrt{2}}{4})$

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6} - \sqrt{2}}{4})$

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6} - \sqrt{2}}{4})$

Step 6

Cancel the common factor of $4$ .

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Step 6.1

Cancel the common factor.

$x = \sqrt{6} + \sqrt{2}$

$y = 4 (\frac{\sqrt{6} - \sqrt{2}}{4})$

Step 6.2

Rewrite the expression.

$x = \sqrt{6} + \sqrt{2}$

$y = \sqrt{6} - \sqrt{2}$

$x = \sqrt{6} + \sqrt{2}$

$y = \sqrt{6} - \sqrt{2}$

Step 7

The rectangular representation of the polar point $(4, \frac{π}{12})$ is $(\sqrt{6} + \sqrt{2}, \sqrt{6} - \sqrt{2})$ .

$(\sqrt{6} + \sqrt{2}, \sqrt{6} - \sqrt{2})$