Precalculus Examples

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

Step 1

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

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Step 1.1

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

$\sin (\frac{\frac{7 π}{6}}{2}) \sin (\frac{π}{12})$

Step 1.2

Apply the sine half-angle identity.

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

Step 1.3

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

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

Step 1.4

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

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Step 1.4.1

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.

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

Step 1.4.2

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

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

Step 1.4.3

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

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Step 1.4.3.1

Multiply $- 1$ by $- 1$ .

$\sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{2}} \sin (\frac{π}{12})$

Step 1.4.3.2

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

$\sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} \sin (\frac{π}{12})$

Step 1.4.4

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

$\sqrt{\frac{\frac{2}{2} + \frac{\sqrt{3}}{2}}{2}} \sin (\frac{π}{12})$

Step 1.4.5

Combine the numerators over the common denominator.

$\sqrt{\frac{\frac{2 + \sqrt{3}}{2}}{2}} \sin (\frac{π}{12})$

Step 1.4.6

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.4.7

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

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Step 1.4.7.1

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

$\sqrt{\frac{2 + \sqrt{3}}{2 \cdot 2}} \sin (\frac{π}{12})$

Step 1.4.7.2

Multiply $2$ by $2$ .

$\sqrt{\frac{2 + \sqrt{3}}{4}} \sin (\frac{π}{12})$

Step 1.4.8

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

$\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{4}} \sin (\frac{π}{12})$

Step 1.4.9

Simplify the denominator.

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Step 1.4.9.1

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

$\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{2^{2}}} \sin (\frac{π}{12})$

Step 1.4.9.2

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

$\frac{\sqrt{2 + \sqrt{3}}}{2} \sin (\frac{π}{12})$

Step 2

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

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Step 2.1

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

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

Step 2.2

Apply the difference of angles identity.

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

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

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

Step 2.7

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

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Step 2.7.1

Simplify each term.

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Step 2.7.1.1

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

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Step 2.7.1.1.1

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

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

Step 2.7.1.1.2

Combine using the product rule for radicals.

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

Step 2.7.1.1.3

Multiply $2$ by $3$ .

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

Step 2.7.1.1.4

Multiply $2$ by $2$ .

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

Step 2.7.1.2

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

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Step 2.7.1.2.1

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

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

Step 2.7.1.2.2

Multiply $2$ by $2$ .

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

Step 2.7.2

Combine the numerators over the common denominator.

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

Step 3

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

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Step 3.1

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

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

Step 3.2

Multiply $2$ by $4$ .

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

Step 4

Apply the distributive property.

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

Step 5

Combine using the product rule for radicals.

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

Step 6

Combine using the product rule for radicals.

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

Step 7

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$0.25$