Trigonometry Examples

\sec (\frac{15 π}{8})

$\sec (\frac{15 π}{8})$

Step 1

Replace

\sec (\frac{15 π}{8})

$\sec (\frac{15 π}{8})$ with an equivalent expression

\frac{1}{\cos (\frac{15 π}{8})}

$\frac{1}{\cos (\frac{15 π}{8})}$ using the fundamental identities.

\frac{1}{\cos (\frac{15 π}{8})}

$\frac{1}{\cos (\frac{15 π}{8})}$

Step 2

Simplify.

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Step 2.1

Convert from

\frac{1}{\cos (\frac{15 π}{8})}

$\frac{1}{\cos (\frac{15 π}{8})}$ to

\sec (\frac{15 π}{8})

$\sec (\frac{15 π}{8})$ .

\sec (\frac{15 π}{8})

$\sec (\frac{15 π}{8})$

Step 2.2

The exact value of

\sec (\frac{15 π}{8})

$\sec (\frac{15 π}{8})$ is

\frac{2}{\sqrt{2 + \sqrt{2}}}

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

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Step 2.2.1

Rewrite

\frac{15 π}{8}

$\frac{15 π}{8}$ as an angle where the values of the six trigonometric functions are known divided by

2

$2$ .

\sec (\frac{\frac{15 π}{4}}{2})

$\sec (\frac{\frac{15 π}{4}}{2})$

Step 2.2.2

Apply the reciprocal identity to

\sec (\frac{\frac{15 π}{4}}{2})

$\sec (\frac{\frac{15 π}{4}}{2})$ .

\frac{1}{\cos (\frac{\frac{15 π}{4}}{2})}

$\frac{1}{\cos (\frac{\frac{15 π}{4}}{2})}$

Step 2.2.3

Apply the cosine half-angle identity

\cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + \cos (x)}{2}}

$\cos (\frac{x}{2}) = \pm \sqrt{\frac{1 + \cos (x)}{2}}$ .

\frac{1}{\pm \sqrt{\frac{1 + \cos (\frac{15 π}{4})}{2}}}

$\frac{1}{\pm \sqrt{\frac{1 + \cos (\frac{15 π}{4})}{2}}}$

Step 2.2.4

Change the

\pm

$\pm$ to

+

$+$ because secant is positive in the fourth quadrant.

\frac{1}{\sqrt{\frac{1 + \cos (\frac{15 π}{4})}{2}}}

$\frac{1}{\sqrt{\frac{1 + \cos (\frac{15 π}{4})}{2}}}$

Step 2.2.5

Simplify

\frac{1}{\sqrt{\frac{1 + \cos (\frac{15 π}{4})}{2}}}

$\frac{1}{\sqrt{\frac{1 + \cos (\frac{15 π}{4})}{2}}}$ .

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Step 2.2.5.1

Simplify the numerator.

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Step 2.2.5.1.1

Subtract full rotations of

2 π

$2 π$ until the angle is greater than or equal to

0

$0$ and less than

2 π

$2 π$ .

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

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

Step 2.2.5.1.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

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

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

Step 2.2.5.1.3

The exact value of

\cos (\frac{π}{4})

$\cos (\frac{π}{4})$ is

\frac{\sqrt{2}}{2}

$\frac{\sqrt{2}}{2}$ .

\frac{1}{\sqrt{\frac{1 + \frac{\sqrt{2}}{2}}{2}}}

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

Step 2.2.5.1.4

Write

1

$1$ as a fraction with a common denominator.

\frac{1}{\sqrt{\frac{\frac{2}{2} + \frac{\sqrt{2}}{2}}{2}}}

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

Step 2.2.5.1.5

Combine the numerators over the common denominator.

\frac{1}{\sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}}}

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

\frac{1}{\sqrt{\frac{\frac{2 + \sqrt{2}}{2}}{2}}}

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

Step 2.2.5.2

Simplify the denominator.

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Step 2.2.5.2.1

Multiply the numerator by the reciprocal of the denominator.

\frac{1}{\sqrt{\frac{2 + \sqrt{2}}{2} \cdot \frac{1}{2}}}

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

Step 2.2.5.2.2

Multiply

\frac{2 + \sqrt{2}}{2} \cdot \frac{1}{2}

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

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Step 2.2.5.2.2.1

Multiply

\frac{2 + \sqrt{2}}{2}

$\frac{2 + \sqrt{2}}{2}$ by

\frac{1}{2}

$\frac{1}{2}$ .

\frac{1}{\sqrt{\frac{2 + \sqrt{2}}{2 \cdot 2}}}

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

Step 2.2.5.2.2.2

Multiply

2

$2$ by

2

$2$ .

\frac{1}{\sqrt{\frac{2 + \sqrt{2}}{4}}}

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

\frac{1}{\sqrt{\frac{2 + \sqrt{2}}{4}}}

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

Step 2.2.5.2.3

Rewrite

\sqrt{\frac{2 + \sqrt{2}}{4}}

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

\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{4}}

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

\frac{1}{\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{4}}}

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

Step 2.2.5.2.4

Simplify the denominator.

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Step 2.2.5.2.4.1

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

\frac{1}{\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2^{2}}}}

$\frac{1}{\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2^{2}}}}$

Step 2.2.5.2.4.2

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

\frac{1}{\frac{\sqrt{2 + \sqrt{2}}}{2}}

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

\frac{1}{\frac{\sqrt{2 + \sqrt{2}}}{2}}

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

\frac{1}{\frac{\sqrt{2 + \sqrt{2}}}{2}}

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

Step 2.2.5.3

Multiply the numerator by the reciprocal of the denominator.

1 \frac{2}{\sqrt{2 + \sqrt{2}}}

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

Step 2.2.5.4

Multiply

\frac{2}{\sqrt{2 + \sqrt{2}}}

$\frac{2}{\sqrt{2 + \sqrt{2}}}$ by

1

$1$ .

\frac{2}{\sqrt{2 + \sqrt{2}}}