Trigonometry Examples

$\sin (\frac{π}{12}) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1

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

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Step 1.1

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

$\sin (\frac{π}{4} - \frac{π}{6}) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.2

Apply the difference of angles identity.

$(\sin (\frac{π}{4}) \cos (\frac{π}{6}) - \cos (\frac{π}{4}) \sin (\frac{π}{6})) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.3

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

$(\frac{\sqrt{2}}{2} \cos (\frac{π}{6}) - \cos (\frac{π}{4}) \sin (\frac{π}{6})) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.4

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

$(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \cos (\frac{π}{4}) \sin (\frac{π}{6})) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.5

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

$(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \sin (\frac{π}{6})) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.6

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

$(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.7

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

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Step 1.7.1

Simplify each term.

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Step 1.7.1.1

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

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Step 1.7.1.1.1

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

$(\frac{\sqrt{2} \sqrt{3}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.7.1.1.2

Combine using the product rule for radicals.

$(\frac{\sqrt{2 \cdot 3}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.7.1.1.3

Multiply $2$ by $3$ .

$(\frac{\sqrt{6}}{2 \cdot 2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.7.1.1.4

Multiply $2$ by $2$ .

$(\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.7.1.2

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

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Step 1.7.1.2.1

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

$(\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{2 \cdot 2}) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.7.1.2.2

Multiply $2$ by $2$ .

$(\frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}) \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 1.7.2

Combine the numerators over the common denominator.

$\frac{\sqrt{6} - \sqrt{2}}{4} \cos (\frac{7 π}{12}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2

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

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Step 2.1

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

$\frac{\sqrt{6} - \sqrt{2}}{4} \cos (\frac{\frac{7 π}{6}}{2}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.2

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

$\frac{\sqrt{6} - \sqrt{2}}{4} (\pm \sqrt{\frac{1 + \cos (\frac{7 π}{6})}{2}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.3

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

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \sqrt{\frac{1 + \cos (\frac{7 π}{6})}{2}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.4

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

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Step 2.4.1

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.

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \sqrt{\frac{1 - \cos (\frac{π}{6})}{2}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.4.2

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

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.4.3

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

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \sqrt{\frac{\frac{2}{2} - \frac{\sqrt{3}}{2}}{2}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.4.4

Combine the numerators over the common denominator.

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.4.5

Multiply the numerator by the reciprocal of the denominator.

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \sqrt{\frac{2 - \sqrt{3}}{2} \cdot \frac{1}{2}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.4.6

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

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Step 2.4.6.1

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

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \sqrt{\frac{2 - \sqrt{3}}{2 \cdot 2}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.4.6.2

Multiply $2$ by $2$ .

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \sqrt{\frac{2 - \sqrt{3}}{4}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.4.7

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

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{4}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.4.8

Simplify the denominator.

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Step 2.4.8.1

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

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2^{2}}}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 2.4.8.2

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

$\frac{\sqrt{6} - \sqrt{2}}{4} (- \frac{\sqrt{2 - \sqrt{3}}}{2}) - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 3

Multiply $\frac{\sqrt{6} - \sqrt{2}}{4} (- \frac{\sqrt{2 - \sqrt{3}}}{2})$ .

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Step 3.1

Multiply $\frac{\sqrt{6} - \sqrt{2}}{4}$ by $\frac{\sqrt{2 - \sqrt{3}}}{2}$ .

$- \frac{(\sqrt{6} - \sqrt{2}) \sqrt{2 - \sqrt{3}}}{4 \cdot 2} - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 3.2

Multiply $4$ by $2$ .

$- \frac{(\sqrt{6} - \sqrt{2}) \sqrt{2 - \sqrt{3}}}{8} - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 4

Apply the distributive property.

$- \frac{\sqrt{6} \sqrt{2 - \sqrt{3}} - \sqrt{2} \sqrt{2 - \sqrt{3}}}{8} - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 5

Combine using the product rule for radicals.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{2} \sqrt{2 - \sqrt{3}}}{8} - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 6

Combine using the product rule for radicals.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \cos (\frac{π}{12}) \sin (\frac{7 π}{12})$

Step 7

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

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Step 7.1

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \cos (\frac{π}{4} - \frac{π}{6}) \sin (\frac{7 π}{12})$

Step 7.2

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\cos (\frac{π}{4}) \cos (\frac{π}{6}) + \sin (\frac{π}{4}) \sin (\frac{π}{6})) \sin (\frac{7 π}{12})$

Step 7.3

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\frac{\sqrt{2}}{2} \cos (\frac{π}{6}) + \sin (\frac{π}{4}) \sin (\frac{π}{6})) \sin (\frac{7 π}{12})$

Step 7.4

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \sin (\frac{π}{4}) \sin (\frac{π}{6})) \sin (\frac{7 π}{12})$

Step 7.5

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \sin (\frac{π}{6})) \sin (\frac{7 π}{12})$

Step 7.6

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \sin (\frac{7 π}{12})$

Step 7.7

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

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Step 7.7.1

Simplify each term.

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Step 7.7.1.1

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

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Step 7.7.1.1.1

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\frac{\sqrt{2} \sqrt{3}}{2 \cdot 2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \sin (\frac{7 π}{12})$

Step 7.7.1.1.2

Combine using the product rule for radicals.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\frac{\sqrt{2 \cdot 3}}{2 \cdot 2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \sin (\frac{7 π}{12})$

Step 7.7.1.1.3

Multiply $2$ by $3$ .

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\frac{\sqrt{6}}{2 \cdot 2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \sin (\frac{7 π}{12})$

Step 7.7.1.1.4

Multiply $2$ by $2$ .

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}) \sin (\frac{7 π}{12})$

Step 7.7.1.2

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

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Step 7.7.1.2.1

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{2 \cdot 2}) \sin (\frac{7 π}{12})$

Step 7.7.1.2.2

Multiply $2$ by $2$ .

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - (\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}) \sin (\frac{7 π}{12})$

Step 7.7.2

Combine the numerators over the common denominator.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sin (\frac{7 π}{12})$

Step 8

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

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Step 8.1

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sin (\frac{\frac{7 π}{6}}{2})$

Step 8.2

Apply the sine half-angle identity.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} (\pm \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{2}})$

Step 8.3

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sqrt{\frac{1 - \cos (\frac{7 π}{6})}{2}}$

Step 8.4

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

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Step 8.4.1

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.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sqrt{\frac{1 - - \cos (\frac{π}{6})}{2}}$

Step 8.4.2

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sqrt{\frac{1 - - \frac{\sqrt{3}}{2}}{2}}$

Step 8.4.3

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

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Step 8.4.3.1

Multiply $- 1$ by $- 1$ .

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{2}}$

Step 8.4.3.2

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}$

Step 8.4.4

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{2}}$

Step 8.4.5

Combine the numerators over the common denominator.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{2}}$

Step 8.4.6

Multiply the numerator by the reciprocal of the denominator.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sqrt{\frac{2 + \sqrt{3}}{2} \cdot \frac{1}{2}}$

Step 8.4.7

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

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Step 8.4.7.1

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sqrt{\frac{2 + \sqrt{3}}{2 \cdot 2}}$

Step 8.4.7.2

Multiply $2$ by $2$ .

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \sqrt{\frac{2 + \sqrt{3}}{4}}$

Step 8.4.8

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \cdot \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}}$

Step 8.4.9

Simplify the denominator.

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Step 8.4.9.1

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \cdot \frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2^{2}}}$

Step 8.4.9.2

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

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{6} + \sqrt{2}}{4} \cdot \frac{\sqrt{2 + \sqrt{3}}}{2}$

Step 9

Multiply $- \frac{\sqrt{6} + \sqrt{2}}{4} \cdot \frac{\sqrt{2 + \sqrt{3}}}{2}$ .

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Step 9.1

Multiply $\frac{\sqrt{2 + \sqrt{3}}}{2}$ by $\frac{\sqrt{6} + \sqrt{2}}{4}$ .

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{2 + \sqrt{3}} (\sqrt{6} + \sqrt{2})}{2 \cdot 4}$

Step 9.2

Multiply $2$ by $4$ .

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{2 + \sqrt{3}} (\sqrt{6} + \sqrt{2})}{8}$

Step 10

Apply the distributive property.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{2 + \sqrt{3}} \sqrt{6} + \sqrt{2 + \sqrt{3}} \sqrt{2}}{8}$

Step 11

Combine using the product rule for radicals.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{(2 + \sqrt{3}) \cdot 6} + \sqrt{2 + \sqrt{3}} \sqrt{2}}{8}$

Step 12

Combine using the product rule for radicals.

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{(2 + \sqrt{3}) \cdot 6} + \sqrt{(2 + \sqrt{3}) \cdot 2}}{8}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$- \frac{\sqrt{6 (2 - \sqrt{3})} - \sqrt{(2 - \sqrt{3}) \cdot 2}}{8} - \frac{\sqrt{(2 + \sqrt{3}) \cdot 6} + \sqrt{(2 + \sqrt{3}) \cdot 2}}{8}$

Decimal Form:

$- 1$