Precalculus Examples

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

Step 1

Simplify each term.

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.1.3

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

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

Step 1.1.4

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

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

Step 1.1.5

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

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

Step 1.1.6

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

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

Step 1.1.7

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

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Step 1.1.7.1

Simplify each term.

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Step 1.1.7.1.1

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

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Step 1.1.7.1.1.1

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

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

Step 1.1.7.1.1.2

Combine using the product rule for radicals.

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

Step 1.1.7.1.1.3

Multiply $2$ by $3$ .

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

Step 1.1.7.1.1.4

Multiply $2$ by $2$ .

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

Step 1.1.7.1.2

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

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Step 1.1.7.1.2.1

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

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

Step 1.1.7.1.2.2

Multiply $2$ by $2$ .

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

Step 1.1.7.2

Combine the numerators over the common denominator.

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

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 $2$ .

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

Step 1.2.2

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

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

Step 1.2.3

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

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

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.

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

Step 1.2.4.2

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

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

Step 1.2.4.3

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

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

Step 1.2.4.4

Combine the numerators over the common denominator.

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

Step 1.2.4.5

Multiply the numerator by the reciprocal of the denominator.

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

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 $\frac{1}{2}$ .

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

Step 1.2.4.6.2

Multiply $2$ by $2$ .

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

Step 1.2.4.7

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

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

Step 1.2.4.8

Simplify the denominator.

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Step 1.2.4.8.1

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

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

Step 1.2.4.8.2

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

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

Step 2

To write $- \frac{\sqrt{2 - \sqrt{3}}}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.1

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

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

Step 3.2

Multiply $2$ by $2$ .

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

Step 4

Combine the numerators over the common denominator.

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

Step 5

Multiply $2$ by $- 1$ .

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$0.70710678 \dots$