Trigonometry Examples

y = 5 \csc (x)

$y = 5 \csc (x)$

Step 1

Find the asymptotes.

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Step 1.1

For any

y = \csc (x)

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

x = n π

$x = n π$ , where

n

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

y = c s c (x)

$y = c s c (x)$ ,

(0, 2 π)

$(0, 2 π)$ , to find the vertical asymptotes for

y = 5 \csc (x)

$y = 5 \csc (x)$ . Set the inside of the cosecant function,

b x + c

$b x + c$ , for

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

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

0

$0$ to find where the vertical asymptote occurs for

y = 5 \csc (x)

$y = 5 \csc (x)$ .

x = 0

$x = 0$

Step 1.2

Set the inside of the cosecant function

x

$x$ equal to

2 π

$2 π$ .

x = 2 π

$x = 2 π$

Step 1.3

The basic period for

y = 5 \csc (x)

$y = 5 \csc (x)$ will occur at

(0, 2 π)

$(0, 2 π)$ , where

0

$0$ and

2 π

$2 π$ are vertical asymptotes.

(0, 2 π)

$(0, 2 π)$

Step 1.4

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 1.4.1

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 1.4.2

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 1.5

The vertical asymptotes for

y = 5 \csc (x)

$y = 5 \csc (x)$ occur at

0

$0$ ,

2 π

$2 π$ , and every

π n

$π n$ , where

n

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

π n

$π n$

Step 1.6

There are only vertical asymptotes for secant and cosecant functions.

Vertical Asymptotes:

x = π n

$x = π n$ for any integer

n

$n$

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes:

x = π n

$x = π n$ for any integer

n

$n$

No Horizontal Asymptotes

No Oblique Asymptotes

Step 2

Use the form

a \csc (b x - c) + d

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

a = 5

$a = 5$

b = 1

$b = 1$

c = 0

$c = 0$

d = 0

$d = 0$

Step 3

Since the graph of the function

c s c

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

Amplitude: None

Step 4

Find the period of

5 \csc (x)

$5 \csc (x)$ .

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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

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 4.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 4.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

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{0}{1}

$\frac{0}{1}$

Step 5.3

Divide

0

$0$ by

1

$1$ .

Phase Shift:

0

$0$

Phase Shift:

0

$0$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period:

2 π

$2 π$

Phase Shift: None

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 = π n

$x = π n$ for any integer

n

$n$

Amplitude: None

Period:

2 π

$2 π$

Phase Shift: None

Vertical Shift: None

Step 8