Precalculus Examples

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

Step 1

Simplify each term.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

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

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Step 1.2.4.1

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.

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

Step 1.2.4.2

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

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

Step 1.2.4.3

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

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

Step 1.2.4.4

Combine the numerators over the common denominator.

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

Step 1.2.4.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.2.4.6

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

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Step 1.2.4.6.1

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

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

Step 1.2.4.6.2

Multiply $2$ by $2$ .

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

Step 1.2.4.7

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

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

Step 1.2.4.8

Simplify the denominator.

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Step 1.2.4.8.1

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

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

Step 1.2.4.8.2

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

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

Step 1.3

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

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Step 1.3.1

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

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

Step 1.3.2

Multiply $2$ by $2$ .

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

Step 1.4

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

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

Step 1.5

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

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Step 1.5.1

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

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

Step 1.5.2

Apply the sine half-angle identity.

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

Step 1.5.3

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

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

Step 1.5.4

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

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Step 1.5.4.1

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.

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

Step 1.5.4.2

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

$- \frac{\sqrt{2 - \sqrt{3}}}{4} + \frac{\sqrt{3}}{2} \sqrt{\frac{1 - - \frac{\sqrt{3}}{2}}{2}}$

Step 1.5.4.3

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

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Step 1.5.4.3.1

Multiply $- 1$ by $- 1$ .

$- \frac{\sqrt{2 - \sqrt{3}}}{4} + \frac{\sqrt{3}}{2} \sqrt{\frac{1 + 1 \frac{\sqrt{3}}{2}}{2}}$

Step 1.5.4.3.2

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

$- \frac{\sqrt{2 - \sqrt{3}}}{4} + \frac{\sqrt{3}}{2} \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}}$

Step 1.5.4.4

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

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

Step 1.5.4.5

Combine the numerators over the common denominator.

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

Step 1.5.4.6

Multiply the numerator by the reciprocal of the denominator.

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

Step 1.5.4.7

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

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Step 1.5.4.7.1

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

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

Step 1.5.4.7.2

Multiply $2$ by $2$ .

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

Step 1.5.4.8

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

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

Step 1.5.4.9

Simplify the denominator.

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Step 1.5.4.9.1

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

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

Step 1.5.4.9.2

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

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

Step 1.6

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

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Step 1.6.1

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

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

Step 1.6.2

Combine using the product rule for radicals.

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

Step 1.6.3

Multiply $2$ by $2$ .

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

Step 2

Simplify with factoring out.

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Step 2.1

Combine the numerators over the common denominator.

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

Step 2.2

Factor $- 1$ out of $- \sqrt{2 - \sqrt{3}}$ .

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

Step 2.3

Factor $- 1$ out of $\sqrt{3 (2 + \sqrt{3})}$ .

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

Step 2.4

Factor $- 1$ out of $- (\sqrt{2 - \sqrt{3}}) - 1 (- \sqrt{3 (2 + \sqrt{3})})$ .

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

Step 2.5

Simplify the expression.

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Step 2.5.1

Rewrite $- (\sqrt{2 - \sqrt{3}} - \sqrt{3 (2 + \sqrt{3})})$ as $- 1 (\sqrt{2 - \sqrt{3}} - \sqrt{3 (2 + \sqrt{3})})$ .

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

Step 2.5.2

Move the negative in front of the fraction.

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

Step 3

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$0.70710678 \dots$