Precalculus Examples

$\cos (\frac{5 π}{8}) = - v \frac{2 - \sqrt{2}}{2}$ , $\sin (\frac{5 π}{16})$

Step 1

Use the definition of cosine to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values.

$\cos (\frac{5 π}{8}) = \frac{adjacent}{hypotenuse}$

Step 2

Find the opposite side of the unit circle triangle. Since the adjacent side and hypotenuse are known, use the Pythagorean theorem to find the remaining side.

$Opposite = \sqrt{{hypotenuse}^{2} - {adjacent}^{2}}$

Step 3

Replace the known values in the equation.

$Opposite = \sqrt{{(1)}^{2} - {(- v \frac{2 - \sqrt{2}}{2})}^{2}}$

Step 4

Simplify inside the radical.

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Step 4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = - v \frac{2 - \sqrt{2}}{2}$ .

Opposite $= \sqrt{(1 - v \frac{2 - \sqrt{2}}{2}) (1 - (- v \frac{2 - \sqrt{2}}{2}))}$

Step 4.2

Simplify.

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Step 4.2.1

Combine $\frac{2 - \sqrt{2}}{2}$ and $v$ .

Opposite $= \sqrt{(1 - \frac{(2 - \sqrt{2}) v}{2}) (1 - (- v \frac{2 - \sqrt{2}}{2}))}$

Step 4.2.2

Combine $\frac{2 - \sqrt{2}}{2}$ and $v$ .

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

Step 4.2.3

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

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Step 4.2.3.1

Multiply $- 1$ by $- 1$ .

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

Step 4.2.3.2

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

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

Step 5

Use the definition of sine to find the value of $\sin (\frac{5 π}{8})$ .

$\sin (\frac{5 π}{8}) = \frac{opposite}{hypotenuse}$

Step 6

Substitute in the known values.

$\sin (\frac{5 π}{8}) = \frac{\sqrt{(1 - \frac{(2 - \sqrt{2}) v}{2}) (1 + \frac{(2 - \sqrt{2}) v}{2})}}{1}$

Step 7

Simplify the right side.

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Step 7.1

Simplify.

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Step 7.1.1

Divide $\sqrt{(1 - \frac{(2 - \sqrt{2}) v}{2}) (1 + \frac{(2 - \sqrt{2}) v}{2})}$ by $1$ .

$\sin (\frac{5 π}{8}) = \sqrt{(1 - \frac{(2 - \sqrt{2}) v}{2}) (1 + \frac{(2 - \sqrt{2}) v}{2})}$

Step 7.1.2

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

$\sin (\frac{5 π}{8}) = \sqrt{(\frac{2}{2} - \frac{(2 - \sqrt{2}) v}{2}) (1 + \frac{(2 - \sqrt{2}) v}{2})}$

Step 7.2

Combine the numerators over the common denominator.

$\sin (\frac{5 π}{8}) = \sqrt{\frac{2 - (2 - \sqrt{2}) v}{2} \cdot (1 + \frac{(2 - \sqrt{2}) v}{2})}$

Step 7.3

Rewrite $\frac{2 - (2 - \sqrt{2}) v}{2}$ in a factored form.

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Step 7.3.1

Apply the distributive property.

$\sin (\frac{5 π}{8}) = \sqrt{\frac{2 + (- 1 \cdot 2 + \sqrt{2}) v}{2} \cdot (1 + \frac{(2 - \sqrt{2}) v}{2})}$

Step 7.3.2

Multiply $- 1$ by $2$ .

$\sin (\frac{5 π}{8}) = \sqrt{\frac{2 + (- 2 + \sqrt{2}) v}{2} \cdot (1 + \frac{(2 - \sqrt{2}) v}{2})}$

Step 7.3.3

Multiply $- - \sqrt{2}$ .

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Step 7.3.3.1

Multiply $- 1$ by $- 1$ .

$\sin (\frac{5 π}{8}) = \sqrt{\frac{2 + (- 2 + 1 \sqrt{2}) v}{2} \cdot (1 + \frac{(2 - \sqrt{2}) v}{2})}$

Step 7.3.3.2

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

$\sin (\frac{5 π}{8}) = \sqrt{\frac{2 + (- 2 + \sqrt{2}) v}{2} \cdot (1 + \frac{(2 - \sqrt{2}) v}{2})}$

Step 7.3.4

Apply the distributive property.

$\sin (\frac{5 π}{8}) = \sqrt{\frac{2 - 2 v + \sqrt{2} v}{2} \cdot (1 + \frac{(2 - \sqrt{2}) v}{2})}$

Step 7.4

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

$\sin (\frac{5 π}{8}) = \sqrt{\frac{2 - 2 v + \sqrt{2} v}{2} \cdot (\frac{2}{2} + \frac{(2 - \sqrt{2}) v}{2})}$

Step 7.5

Combine the numerators over the common denominator.

$\sin (\frac{5 π}{8}) = \sqrt{\frac{2 - 2 v + \sqrt{2} v}{2} \cdot \frac{2 + (2 - \sqrt{2}) v}{2}}$

Step 7.6

Apply the distributive property.

$\sin (\frac{5 π}{8}) = \sqrt{\frac{2 - 2 v + \sqrt{2} v}{2} \cdot \frac{2 + 2 v - \sqrt{2} v}{2}}$

Step 7.7

Multiply $\frac{2 - 2 v + \sqrt{2} v}{2}$ by $\frac{2 + 2 v - \sqrt{2} v}{2}$ .

$\sin (\frac{5 π}{8}) = \sqrt{\frac{(2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v)}{2 \cdot 2}}$

Step 7.8

Multiply $2$ by $2$ .

$\sin (\frac{5 π}{8}) = \sqrt{\frac{(2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v)}{4}}$

Step 7.9

Rewrite $\frac{(2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v)}{4}$ as ${(\frac{1}{2})}^{2} ((2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v))$ .

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Step 7.9.1

Factor the perfect power $1^{2}$ out of $(2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v)$ .

$\sin (\frac{5 π}{8}) = \sqrt{\frac{1^{2} ((2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v))}{4}}$

Step 7.9.2

Factor the perfect power $2^{2}$ out of $4$ .

$\sin (\frac{5 π}{8}) = \sqrt{\frac{1^{2} ((2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v))}{2^{2} \cdot 1}}$

Step 7.9.3

Rearrange the fraction $\frac{1^{2} ((2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v))}{2^{2} \cdot 1}$ .

$\sin (\frac{5 π}{8}) = \sqrt{{(\frac{1}{2})}^{2} ((2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v))}$

Step 7.10

Pull terms out from under the radical.

$\sin (\frac{5 π}{8}) = \frac{1}{2} \cdot \sqrt{(2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v)}$

Step 7.11

Combine $\frac{1}{2}$ and $\sqrt{(2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v)}$ .

$\sin (\frac{5 π}{8}) = \frac{\sqrt{(2 - 2 v + \sqrt{2} v) (2 + 2 v - \sqrt{2} v)}}{2}$

Step 8

Evaluate $\sin (\frac{5 π}{16})$ .

$0.83146961$

Step 9

Substitute the values into $0.83146961$ .

$\sin (\frac{5 π}{16}) = 0.83146961$