Precalculus Examples

$\cos^{2} (\frac{π}{8}) - \sin^{2} (\frac{π}{8})$

Step 1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = \cos (\frac{π}{8})$ and $b = \sin (\frac{π}{8})$ .

$(\cos (\frac{π}{8}) + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2

Simplify.

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Step 2.1

The exact value of $\cos (\frac{π}{8})$ is $\frac{\sqrt{2 + \sqrt{2}}}{2}$ .

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Step 2.1.1

Rewrite $\frac{π}{8}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$(\cos (\frac{\frac{π}{4}}{2}) + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.1.2

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

$(\pm \sqrt{\frac{1 + \cos (\frac{π}{4})}{2}} + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.1.3

Change the $\pm$ to $+$ because cosine is positive in the first quadrant.

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

Step 2.1.4

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

$(\sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.1.5

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

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Step 2.1.5.1

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

$(\sqrt{\frac{\frac{2}{2} + \frac{\sqrt{2}}{2}}{2}} + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.1.5.2

Combine the numerators over the common denominator.

$(\sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}} + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.1.5.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.1.5.4

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

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Step 2.1.5.4.1

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

$(\sqrt{\frac{2 + \sqrt{2}}{2 \cdot 2}} + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.1.5.4.2

Multiply $2$ by $2$ .

$(\sqrt{\frac{2 + \sqrt{2}}{4}} + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.1.5.5

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

$(\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{4}} + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.1.5.6

Simplify the denominator.

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Step 2.1.5.6.1

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

$(\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2^{2}}} + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.1.5.6.2

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

$(\frac{\sqrt{2 + \sqrt{2}}}{2} + \sin (\frac{π}{8})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.2

The exact value of $\sin (\frac{π}{8})$ is $\frac{\sqrt{2 - \sqrt{2}}}{2}$ .

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Step 2.2.1

Rewrite $\frac{π}{8}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$(\frac{\sqrt{2 + \sqrt{2}}}{2} + \sin (\frac{\frac{π}{4}}{2})) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.2.2

Apply the sine half-angle identity.

$(\frac{\sqrt{2 + \sqrt{2}}}{2} \pm \sqrt{\frac{1 - \cos (\frac{π}{4})}{2}}) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.2.3

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

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

Step 2.2.4

Simplify $\sqrt{\frac{1 - \cos (\frac{π}{4})}{2}}$ .

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Step 2.2.4.1

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

$(\frac{\sqrt{2 + \sqrt{2}}}{2} + \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}}) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.2.4.2

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

$(\frac{\sqrt{2 + \sqrt{2}}}{2} + \sqrt{\frac{\frac{2}{2} - \frac{\sqrt{2}}{2}}{2}}) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.2.4.3

Combine the numerators over the common denominator.

$(\frac{\sqrt{2 + \sqrt{2}}}{2} + \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}}) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.2.4.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.2.4.5

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

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Step 2.2.4.5.1

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

$(\frac{\sqrt{2 + \sqrt{2}}}{2} + \sqrt{\frac{2 - \sqrt{2}}{2 \cdot 2}}) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.2.4.5.2

Multiply $2$ by $2$ .

$(\frac{\sqrt{2 + \sqrt{2}}}{2} + \sqrt{\frac{2 - \sqrt{2}}{4}}) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.2.4.6

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

$(\frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{4}}) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.2.4.7

Simplify the denominator.

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Step 2.2.4.7.1

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

$(\frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2^{2}}}) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.2.4.7.2

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

$(\frac{\sqrt{2 + \sqrt{2}}}{2} + \frac{\sqrt{2 - \sqrt{2}}}{2}) (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.3

Combine the numerators over the common denominator.

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\cos (\frac{π}{8}) - \sin (\frac{π}{8}))$

Step 2.4

The exact value of $\cos (\frac{π}{8})$ is $\frac{\sqrt{2 + \sqrt{2}}}{2}$ .

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Step 2.4.1

Rewrite $\frac{π}{8}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\cos (\frac{\frac{π}{4}}{2}) - \sin (\frac{π}{8}))$

Step 2.4.2

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

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\pm \sqrt{\frac{1 + \cos (\frac{π}{4})}{2}} - \sin (\frac{π}{8}))$

Step 2.4.3

Change the $\pm$ to $+$ because cosine is positive in the first quadrant.

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

Step 2.4.4

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

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}} - \sin (\frac{π}{8}))$

Step 2.4.5

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

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Step 2.4.5.1

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

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\sqrt{\frac{\frac{2}{2} + \frac{\sqrt{2}}{2}}{2}} - \sin (\frac{π}{8}))$

Step 2.4.5.2

Combine the numerators over the common denominator.

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}} - \sin (\frac{π}{8}))$

Step 2.4.5.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.4.5.4

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

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Step 2.4.5.4.1

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

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\sqrt{\frac{2 + \sqrt{2}}{2 \cdot 2}} - \sin (\frac{π}{8}))$

Step 2.4.5.4.2

Multiply $2$ by $2$ .

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\sqrt{\frac{2 + \sqrt{2}}{4}} - \sin (\frac{π}{8}))$

Step 2.4.5.5

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

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{4}} - \sin (\frac{π}{8}))$

Step 2.4.5.6

Simplify the denominator.

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Step 2.4.5.6.1

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

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2^{2}}} - \sin (\frac{π}{8}))$

Step 2.4.5.6.2

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

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\frac{\sqrt{2 + \sqrt{2}}}{2} - \sin (\frac{π}{8}))$

Step 2.5

The exact value of $\sin (\frac{π}{8})$ is $\frac{\sqrt{2 - \sqrt{2}}}{2}$ .

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Step 2.5.1

Rewrite $\frac{π}{8}$ as an angle where the values of the six trigonometric functions are known divided by $2$ .

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

Step 2.5.2

Apply the sine half-angle identity.

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

Step 2.5.3

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

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

Step 2.5.4

Simplify $\sqrt{\frac{1 - \cos (\frac{π}{4})}{2}}$ .

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Step 2.5.4.1

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

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

Step 2.5.4.2

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

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

Step 2.5.4.3

Combine the numerators over the common denominator.

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

Step 2.5.4.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 2.5.4.5

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

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Step 2.5.4.5.1

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

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

Step 2.5.4.5.2

Multiply $2$ by $2$ .

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

Step 2.5.4.6

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

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

Step 2.5.4.7

Simplify the denominator.

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Step 2.5.4.7.1

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

$\frac{\sqrt{2 + \sqrt{2}} + \sqrt{2 - \sqrt{2}}}{2} (\frac{\sqrt{2 + \sqrt{2}}}{2} - \frac{\sqrt{2 - \sqrt{2}}}{\sqrt{2^{2}}})$

Step 2.5.4.7.2

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

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

Step 2.6

Combine the numerators over the common denominator.

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

Step 3

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$0.70710678 \dots$