Precalculus Examples

$(- 1, \frac{π}{8})$

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 $r = - 1$ and $θ = \frac{π}{8}$ into the formulas.

$x = (- 1) \cos (\frac{π}{8})$

$y = (- 1) \sin (\frac{π}{8})$

Step 3

The exact value of $\cos (\frac{π}{8})$ is $\frac{\sqrt{2 + \sqrt{2}}}{2}$ .

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Step 3.1

Rewrite $\frac{π}{8}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$x = - 1 \cos (\frac{\frac{π}{4}}{2})$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.2

Apply the cosine half-angle identity $\cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + \cos (x)}{2}}$ .

$x = - 1 (\pm \sqrt{\frac{1 + \cos (\frac{π}{4})}{2}})$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.3

Change the $\pm$ to $+$ because cosine is positive in the first quadrant.

$x = - 1 \sqrt{\frac{1 + \cos (\frac{π}{4})}{2}}$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.4

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

$x = - 1 \sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.5

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

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Step 3.5.1

Write $1$ as a fraction with a common denominator.

$x = - 1 \sqrt{\frac{\frac{2}{2} + \frac{\sqrt{2}}{2}}{2}}$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.5.2

Combine the numerators over the common denominator.

$x = - 1 \sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}}$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.5.3

Multiply the numerator by the reciprocal of the denominator.

$x = - 1 \sqrt{\frac{2 + \sqrt{2}}{2} \cdot \frac{1}{2}}$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.5.4

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

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Step 3.5.4.1

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

$x = - 1 \sqrt{\frac{2 + \sqrt{2}}{2 \cdot 2}}$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.5.4.2

Multiply $2$ by $2$ .

$x = - 1 \sqrt{\frac{2 + \sqrt{2}}{4}}$

$y = (- 1) \sin (\frac{π}{8})$

$x = - 1 \sqrt{\frac{2 + \sqrt{2}}{4}}$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.5.5

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

$x = - 1 \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{4}}$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.5.6

Simplify the denominator.

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Step 3.5.6.1

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

$x = - 1 \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2^{2}}}$

$y = (- 1) \sin (\frac{π}{8})$

Step 3.5.6.2

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

$x = - 1 \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = (- 1) \sin (\frac{π}{8})$

$x = - 1 \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = (- 1) \sin (\frac{π}{8})$

$x = - 1 \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = (- 1) \sin (\frac{π}{8})$

$x = - 1 \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = (- 1) \sin (\frac{π}{8})$

Step 4

Rewrite $- 1 \frac{\sqrt{2 + \sqrt{2}}}{2}$ as $- \frac{\sqrt{2 + \sqrt{2}}}{2}$ .

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = (- 1) \sin (\frac{π}{8})$

Step 5

The exact value of $\sin (\frac{π}{8})$ is $\frac{\sqrt{2 - \sqrt{2}}}{2}$ .

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Step 5.1

Rewrite $\frac{π}{8}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \sin (\frac{\frac{π}{4}}{2})$

Step 5.2

Apply the sine half-angle identity.

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 (\pm \sqrt{\frac{1 - \cos (\frac{π}{4})}{2}})$

Step 5.3

Change the $\pm$ to $+$ because sine is positive in the first quadrant.

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \sqrt{\frac{1 - \cos (\frac{π}{4})}{2}}$

Step 5.4

Simplify $\sqrt{\frac{1 - \cos (\frac{π}{4})}{2}}$ .

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Step 5.4.1

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

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}}$

Step 5.4.2

Write $1$ as a fraction with a common denominator.

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \sqrt{\frac{\frac{2}{2} - \frac{\sqrt{2}}{2}}{2}}$

Step 5.4.3

Combine the numerators over the common denominator.

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}}$

Step 5.4.4

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \sqrt{\frac{2 - \sqrt{2}}{2} \cdot \frac{1}{2}}$

Step 5.4.5

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

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Step 5.4.5.1

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

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \sqrt{\frac{2 - \sqrt{2}}{2 \cdot 2}}$

Step 5.4.5.2

Multiply $2$ by $2$ .

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \sqrt{\frac{2 - \sqrt{2}}{4}}$

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \sqrt{\frac{2 - \sqrt{2}}{4}}$

Step 5.4.6

Rewrite $\sqrt{\frac{2 - \sqrt{2}}{4}}$ as $\frac{\sqrt{2 - \sqrt{2}}}{\sqrt{4}}$ .

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{4}}$

Step 5.4.7

Simplify the denominator.

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Step 5.4.7.1

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

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2^{2}}}$

Step 5.4.7.2

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

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \frac{\sqrt{2 - \sqrt{2}}}{2}$

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \frac{\sqrt{2 - \sqrt{2}}}{2}$

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \frac{\sqrt{2 - \sqrt{2}}}{2}$

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

$y = - 1 \frac{\sqrt{2 - \sqrt{2}}}{2}$

Step 6

Rewrite $- 1 \frac{\sqrt{2 - \sqrt{2}}}{2}$ as $- \frac{\sqrt{2 - \sqrt{2}}}{2}$ .

$x = - \frac{\sqrt{2 + \sqrt{2}}}{2}$

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

Step 7

The rectangular representation of the polar point $(- 1, \frac{π}{8})$ is $(- \frac{\sqrt{2 + \sqrt{2}}}{2}, - \frac{\sqrt{2 - \sqrt{2}}}{2})$ .

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