Trigonometry Examples

y = \sec (2 x + \frac{π}{2})

$y = \sec (2 x + \frac{π}{2})$

Step 1

Find the asymptotes.

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Step 1.1

For any

y = \sec (x)

$y = \sec (x)$ , vertical asymptotes occur at

x = \frac{π}{2} + n π

$x = \frac{π}{2} + n π$ , where

n

$n$ is an integer. Use the basic period for

y = s e c (x)

$y = s e c (x)$ ,

(- \frac{π}{2}, \frac{3 π}{2})

$(- \frac{π}{2}, \frac{3 π}{2})$ , to find the vertical asymptotes for

y = \sec (2 x + \frac{π}{2})

$y = \sec (2 x + \frac{π}{2})$ . Set the inside of the secant function,

b x + c

$b x + c$ , for

y = a \sec (b x + c) + d

$y = a \sec (b x + c) + d$ equal to

- \frac{π}{2}

$- \frac{π}{2}$ to find where the vertical asymptote occurs for

y = \sec (2 x + \frac{π}{2})

$y = \sec (2 x + \frac{π}{2})$ .

2 x + \frac{π}{2} = - \frac{π}{2}

$2 x + \frac{π}{2} = - \frac{π}{2}$

Step 1.2

Solve for

x

$x$ .

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Step 1.2.1

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 1.2.1.1

Subtract

\frac{π}{2}

$\frac{π}{2}$ from both sides of the equation.

2 x = - \frac{π}{2} - \frac{π}{2}

$2 x = - \frac{π}{2} - \frac{π}{2}$

Step 1.2.1.2

Combine the numerators over the common denominator.

2 x = \frac{- π - π}{2}

$2 x = \frac{- π - π}{2}$

Step 1.2.1.3

Subtract

π

$π$ from

- π

$- π$ .

2 x = \frac{- 2 π}{2}

$2 x = \frac{- 2 π}{2}$

Step 1.2.1.4

Cancel the common factor of

- 2

$- 2$ and

2

$2$ .

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Step 1.2.1.4.1

Factor

2

$2$ out of

- 2 π

$- 2 π$ .

2 x = \frac{2 (- π)}{2}

$2 x = \frac{2 (- π)}{2}$

Step 1.2.1.4.2

Cancel the common factors.

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Step 1.2.1.4.2.1

Factor

2

$2$ out of

2

$2$ .

2 x = \frac{2 (- π)}{2 (1)}

$2 x = \frac{2 (- π)}{2 (1)}$

Step 1.2.1.4.2.2

Cancel the common factor.

2 x = \frac{2 (- π)}{2 \cdot 1}

$2 x = \frac{2 (- π)}{2 \cdot 1}$

Step 1.2.1.4.2.3

Rewrite the expression.

2 x = \frac{- π}{1}

$2 x = \frac{- π}{1}$

Step 1.2.1.4.2.4

Divide

- π

$- π$ by

1

$1$ .

2 x = - π

$2 x = - π$

2 x = - π

$2 x = - π$

2 x = - π

$2 x = - π$

2 x = - π

$2 x = - π$

Step 1.2.2

Divide each term in

2 x = - π

$2 x = - π$ by

2

$2$ and simplify.

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Step 1.2.2.1

Divide each term in

2 x = - π

$2 x = - π$ by

2

$2$ .

\frac{2 x}{2} = \frac{- π}{2}

$\frac{2 x}{2} = \frac{- π}{2}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of

2

$2$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

\frac{2 x}{2} = \frac{- π}{2}

$\frac{2 x}{2} = \frac{- π}{2}$

Step 1.2.2.2.1.2

Divide

x

$x$ by

1

$1$ .

x = \frac{- π}{2}

$x = \frac{- π}{2}$

x = \frac{- π}{2}

$x = \frac{- π}{2}$

x = \frac{- π}{2}

$x = \frac{- π}{2}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Move the negative in front of the fraction.

x = - \frac{π}{2}

$x = - \frac{π}{2}$

x = - \frac{π}{2}

$x = - \frac{π}{2}$

x = - \frac{π}{2}

$x = - \frac{π}{2}$

x = - \frac{π}{2}

$x = - \frac{π}{2}$

Step 1.3

Set the inside of the secant function

2 x + \frac{π}{2}

$2 x + \frac{π}{2}$ equal to

\frac{3 π}{2}

$\frac{3 π}{2}$ .

2 x + \frac{π}{2} = \frac{3 π}{2}

$2 x + \frac{π}{2} = \frac{3 π}{2}$

Step 1.4

Solve for

x

$x$ .

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Step 1.4.1

Move all terms not containing

x

$x$ to the right side of the equation.

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Step 1.4.1.1

Subtract

\frac{π}{2}

$\frac{π}{2}$ from both sides of the equation.

2 x = \frac{3 π}{2} - \frac{π}{2}

$2 x = \frac{3 π}{2} - \frac{π}{2}$

Step 1.4.1.2

Combine the numerators over the common denominator.

2 x = \frac{3 π - π}{2}

$2 x = \frac{3 π - π}{2}$

Step 1.4.1.3

Subtract

π

$π$ from

3 π

$3 π$ .

2 x = \frac{2 π}{2}

$2 x = \frac{2 π}{2}$

Step 1.4.1.4

Cancel the common factor of

2

$2$ .

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Step 1.4.1.4.1

Cancel the common factor.

2 x = \frac{2 π}{2}

$2 x = \frac{2 π}{2}$

Step 1.4.1.4.2

Divide

π

$π$ by

1

$1$ .

2 x = π

$2 x = π$

2 x = π

$2 x = π$

2 x = π

$2 x = π$

Step 1.4.2

Divide each term in

2 x = π

$2 x = π$ by

2

$2$ and simplify.

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Step 1.4.2.1

Divide each term in

2 x = π

$2 x = π$ by

2

$2$ .

\frac{2 x}{2} = \frac{π}{2}

$\frac{2 x}{2} = \frac{π}{2}$

Step 1.4.2.2

Simplify the left side.

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Step 1.4.2.2.1

Cancel the common factor of

2

$2$ .

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Step 1.4.2.2.1.1

Cancel the common factor.

\frac{2 x}{2} = \frac{π}{2}

$\frac{2 x}{2} = \frac{π}{2}$

Step 1.4.2.2.1.2

Divide

x

$x$ by

1

$1$ .

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

Step 1.5

The basic period for

y = \sec (2 x + \frac{π}{2})

$y = \sec (2 x + \frac{π}{2})$ will occur at

(- \frac{π}{2}, \frac{π}{2})

$(- \frac{π}{2}, \frac{π}{2})$ , where

- \frac{π}{2}

$- \frac{π}{2}$ and

\frac{π}{2}

$\frac{π}{2}$ are vertical asymptotes.

(- \frac{π}{2}, \frac{π}{2})

$(- \frac{π}{2}, \frac{π}{2})$

Step 1.6

Find the period

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ to find where the vertical asymptotes exist. Vertical asymptotes occur every half period.

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Step 1.6.1

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

2

$2$ is

2

$2$ .

\frac{2 π}{2}

$\frac{2 π}{2}$

Step 1.6.2

Cancel the common factor of

2

$2$ .

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Step 1.6.2.1

Cancel the common factor.

\frac{2 π}{2}

$\frac{2 π}{2}$

Step 1.6.2.2

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

π

$π$

Step 1.7

The vertical asymptotes for

y = \sec (2 x + \frac{π}{2})

$y = \sec (2 x + \frac{π}{2})$ occur at

- \frac{π}{2}

$- \frac{π}{2}$ ,

\frac{π}{2}

$\frac{π}{2}$ , and every

x = - \frac{π}{2} + \frac{π n}{2}

$x = - \frac{π}{2} + \frac{π n}{2}$ , where

n

$n$ is an integer. This is half of the period.

x = - \frac{π}{2} + \frac{π n}{2}

$x = - \frac{π}{2} + \frac{π n}{2}$

Step 1.8

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes:

x = - \frac{π}{2} + \frac{π n}{2}

$x = - \frac{π}{2} + \frac{π n}{2}$ where

n

$n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes:

x = - \frac{π}{2} + \frac{π n}{2}

$x = - \frac{π}{2} + \frac{π n}{2}$ where

n

$n$ is an integer

Step 2

Use the form

a \sec (b x - c) + d

$a \sec (b x - c) + d$ to find the variables used to find the amplitude, period, phase shift, and vertical shift.

a = 1

$a = 1$

b = 2

$b = 2$

c = - \frac{π}{2}

$c = - \frac{π}{2}$

d = 0

$d = 0$

Step 3

Since the graph of the function

s e c

$s e c$ does not have a maximum or minimum value, there can be no value for the amplitude.

Amplitude: None

Step 4

Find the period of

\sec (2 x + \frac{π}{2})

$\sec (2 x + \frac{π}{2})$ .

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Step 4.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 4.2

Replace

b

$b$ with

2

$2$ in the formula for period.

\frac{2 π}{| 2 |}

$\frac{2 π}{| 2 |}$

Step 4.3

The absolute value is the distance between a number and zero. The distance between

0

$0$ and

2

$2$ is

2

$2$ .

\frac{2 π}{2}

$\frac{2 π}{2}$

Step 4.4

Cancel the common factor of

2

$2$ .

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Step 4.4.1

Cancel the common factor.

\frac{2 π}{2}

$\frac{2 π}{2}$

Step 4.4.2

Divide

π

$π$ by

1

$1$ .

π

$π$

π

$π$

π

$π$

Step 5

Find the phase shift using the formula

\frac{c}{b}

$\frac{c}{b}$ .

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Step 5.1

The phase shift of the function can be calculated from

\frac{c}{b}

$\frac{c}{b}$ .

Phase Shift:

\frac{c}{b}

$\frac{c}{b}$

Step 5.2

Replace the values of

c

$c$ and

b

$b$ in the equation for phase shift.

Phase Shift:

\frac{- \frac{π}{2}}{2}

$\frac{- \frac{π}{2}}{2}$

Step 5.3

Multiply the numerator by the reciprocal of the denominator.

Phase Shift:

- \frac{π}{2} \cdot \frac{1}{2}

$- \frac{π}{2} \cdot \frac{1}{2}$

Step 5.4

Multiply

- \frac{π}{2} \cdot \frac{1}{2}

$- \frac{π}{2} \cdot \frac{1}{2}$ .

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Step 5.4.1

Multiply

\frac{1}{2}

$\frac{1}{2}$ by

\frac{π}{2}

$\frac{π}{2}$ .

Phase Shift:

- \frac{π}{2 \cdot 2}

$- \frac{π}{2 \cdot 2}$

Step 5.4.2

Multiply

2

$2$ by

2

$2$ .

Phase Shift:

- \frac{π}{4}

$- \frac{π}{4}$

Phase Shift:

- \frac{π}{4}

$- \frac{π}{4}$

Phase Shift:

- \frac{π}{4}

$- \frac{π}{4}$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period:

π

$π$

Phase Shift:

- \frac{π}{4}

$- \frac{π}{4}$ (

\frac{π}{4}

$\frac{π}{4}$ to the left)

Vertical Shift: None

Step 7

The trig function can be graphed using the amplitude, period, phase shift, vertical shift, and the points.

Vertical Asymptotes:

x = - \frac{π}{2} + \frac{π n}{2}

$x = - \frac{π}{2} + \frac{π n}{2}$ where

n

$n$ is an integer

Amplitude: None

Period:

π

$π$

Phase Shift:

- \frac{π}{4}

$- \frac{π}{4}$ (

\frac{π}{4}

$\frac{π}{4}$ to the left)

Vertical Shift: None

Step 8