Precalculus Examples

$(4, \frac{7 π}{12})$

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 $θ = \frac{7 π}{12}$ into the formulas.

$x = (4) \cos (\frac{7 π}{12})$

$y = (4) \sin (\frac{7 π}{12})$

Step 3

The exact value of $\cos (\frac{7 π}{12})$ is $- \frac{\sqrt{2 - \sqrt{3}}}{2}$ .

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Step 3.1

Rewrite $\frac{7 π}{12}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$x = 4 \cos (\frac{\frac{7 π}{6}}{2})$

$y = (4) \sin (\frac{7 π}{12})$

Step 3.2

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

$x = 4 (\pm \sqrt{\frac{1 + \cos (\frac{7 π}{6})}{2}})$

$y = (4) \sin (\frac{7 π}{12})$

Step 3.3

Change the $\pm$ to $-$ because cosine is negative in the second quadrant.

$x = 4 (- \sqrt{\frac{1 + \cos (\frac{7 π}{6})}{2}})$

$y = (4) \sin (\frac{7 π}{12})$

Step 3.4

Simplify $- \sqrt{\frac{1 + \cos (\frac{7 π}{6})}{2}}$ .

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Step 3.4.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the third quadrant.

$x = 4 (- \sqrt{\frac{1 - \cos (\frac{π}{6})}{2}})$

$y = (4) \sin (\frac{7 π}{12})$

Step 3.4.2

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

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

$y = (4) \sin (\frac{7 π}{12})$

Step 3.4.3

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

$x = 4 (- \sqrt{\frac{\frac{2}{2} - \frac{\sqrt{3}}{2}}{2}})$

$y = (4) \sin (\frac{7 π}{12})$

Step 3.4.4

Combine the numerators over the common denominator.

$x = 4 (- \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}})$

$y = (4) \sin (\frac{7 π}{12})$

Step 3.4.5

Multiply the numerator by the reciprocal of the denominator.

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

$y = (4) \sin (\frac{7 π}{12})$

Step 3.4.6

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

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Step 3.4.6.1

Multiply $\frac{2 - \sqrt{3}}{2}$ by $\frac{1}{2}$ .

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

$y = (4) \sin (\frac{7 π}{12})$

Step 3.4.6.2

Multiply $2$ by $2$ .

$x = 4 (- \sqrt{\frac{2 - \sqrt{3}}{4}})$

$y = (4) \sin (\frac{7 π}{12})$

$x = 4 (- \sqrt{\frac{2 - \sqrt{3}}{4}})$

$y = (4) \sin (\frac{7 π}{12})$

Step 3.4.7

Rewrite $\sqrt{\frac{2 - \sqrt{3}}{4}}$ as $\frac{\sqrt{2 - \sqrt{3}}}{\sqrt{4}}$ .

$x = 4 (- \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{4}})$

$y = (4) \sin (\frac{7 π}{12})$

Step 3.4.8

Simplify the denominator.

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Step 3.4.8.1

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

$x = 4 (- \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2^{2}}})$

$y = (4) \sin (\frac{7 π}{12})$

Step 3.4.8.2

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

$x = 4 (- \frac{\sqrt{2 - \sqrt{3}}}{2})$

$y = (4) \sin (\frac{7 π}{12})$

$x = 4 (- \frac{\sqrt{2 - \sqrt{3}}}{2})$

$y = (4) \sin (\frac{7 π}{12})$

$x = 4 (- \frac{\sqrt{2 - \sqrt{3}}}{2})$

$y = (4) \sin (\frac{7 π}{12})$

$x = 4 (- \frac{\sqrt{2 - \sqrt{3}}}{2})$

$y = (4) \sin (\frac{7 π}{12})$

Step 4

Cancel the common factor of $2$ .

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Step 4.1

Move the leading negative in $- \frac{\sqrt{2 - \sqrt{3}}}{2}$ into the numerator.

$x = 4 (\frac{- \sqrt{2 - \sqrt{3}}}{2})$

$y = (4) \sin (\frac{7 π}{12})$

Step 4.2

Factor $2$ out of $4$ .

$x = 2 (2) (\frac{- \sqrt{2 - \sqrt{3}}}{2})$

$y = (4) \sin (\frac{7 π}{12})$

Step 4.3

Cancel the common factor.

$x = 2 \cdot (2 (\frac{- \sqrt{2 - \sqrt{3}}}{2}))$

$y = (4) \sin (\frac{7 π}{12})$

Step 4.4

Rewrite the expression.

$x = 2 (- \sqrt{2 - \sqrt{3}})$

$y = (4) \sin (\frac{7 π}{12})$

$x = 2 (- \sqrt{2 - \sqrt{3}})$

$y = (4) \sin (\frac{7 π}{12})$

Step 5

Multiply $- 1$ by $2$ .

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = (4) \sin (\frac{7 π}{12})$

Step 6

The exact value of $\sin (\frac{7 π}{12})$ is $\frac{\sqrt{2 + \sqrt{3}}}{2}$ .

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Step 6.1

Rewrite $\frac{7 π}{12}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 \sin (\frac{\frac{7 π}{6}}{2})$

Step 6.2

Apply the sine half-angle identity.

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 (\pm \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{2}})$

Step 6.3

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

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{2}}$

Step 6.4

Simplify $\sqrt{\frac{1 - \cos (\frac{7 π}{6})}{2}}$ .

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Step 6.4.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the third quadrant.

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 \sqrt{\frac{1 + \cos (\frac{π}{6})}{2}}$

Step 6.4.2

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

$x = - 2 \sqrt{2 - \sqrt{3}}$

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

Step 6.4.3

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

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Step 6.4.3.1

Multiply $- 1$ by $- 1$ .

$x = - 2 \sqrt{2 - \sqrt{3}}$

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

Step 6.4.3.2

Multiply $\frac{\sqrt{3}}{2}$ by $1$ .

$x = - 2 \sqrt{2 - \sqrt{3}}$

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

$x = - 2 \sqrt{2 - \sqrt{3}}$

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

Step 6.4.4

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

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 \sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{2}}$

Step 6.4.5

Combine the numerators over the common denominator.

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{2}}$

Step 6.4.6

Multiply the numerator by the reciprocal of the denominator.

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 \sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{1}{2}}$

Step 6.4.7

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

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Step 6.4.7.1

Multiply $\frac{2 + \sqrt{3}}{2}$ by $\frac{1}{2}$ .

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 \sqrt{\frac{2 + \sqrt{3}}{2 \cdot 2}}$

Step 6.4.7.2

Multiply $2$ by $2$ .

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 \sqrt{\frac{2 + \sqrt{3}}{4}}$

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 \sqrt{\frac{2 + \sqrt{3}}{4}}$

Step 6.4.8

Rewrite $\sqrt{\frac{2 + \sqrt{3}}{4}}$ as $\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}}$ .

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 (\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}})$

Step 6.4.9

Simplify the denominator.

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Step 6.4.9.1

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

$x = - 2 \sqrt{2 - \sqrt{3}}$

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

Step 6.4.9.2

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

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 (\frac{\sqrt{2 + \sqrt{3}}}{2})$

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 (\frac{\sqrt{2 + \sqrt{3}}}{2})$

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 (\frac{\sqrt{2 + \sqrt{3}}}{2})$

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 4 (\frac{\sqrt{2 + \sqrt{3}}}{2})$

Step 7

Cancel the common factor of $2$ .

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Step 7.1

Factor $2$ out of $4$ .

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 2 (2) (\frac{\sqrt{2 + \sqrt{3}}}{2})$

Step 7.2

Cancel the common factor.

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 2 \cdot (2 (\frac{\sqrt{2 + \sqrt{3}}}{2}))$

Step 7.3

Rewrite the expression.

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 2 \sqrt{2 + \sqrt{3}}$

$x = - 2 \sqrt{2 - \sqrt{3}}$

$y = 2 \sqrt{2 + \sqrt{3}}$

Step 8

The rectangular representation of the polar point $(4, \frac{7 π}{12})$ is $(- 2 \sqrt{2 - \sqrt{3}}, 2 \sqrt{2 + \sqrt{3}})$ .

$(- 2 \sqrt{2 - \sqrt{3}}, 2 \sqrt{2 + \sqrt{3}})$