Trigonometry Examples

$f (x) = 3 \csc (2 x)$

Step 1

Find the asymptotes.

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For any $y = \csc (x)$ , vertical asymptotes occur at $x = n π$ , where $n$ is an integer. Use the basic period for $y = c s c (x)$ , $(0, 2 π)$ , to find the vertical asymptotes for $y = 3 \csc (2 x)$ . Set the inside of the cosecant function, $b x + c$ , for $y = a \csc (b x + c) + d$ equal to $0$ to find where the vertical asymptote occurs for $y = 3 \csc (2 x)$ .

$2 x = 0$

Divide each term in $2 x = 0$ by $2$ and simplify.

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Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Simplify the right side.

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Divide $0$ by $2$ .

$x = 0$

Set the inside of the cosecant function $2 x$ equal to $2 π$ .

$2 x = 2 π$

Divide each term in $2 x = 2 π$ by $2$ and simplify.

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Divide each term in $2 x = 2 π$ by $2$ .

$\frac{2 x}{2} = \frac{2 π}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 x}{2} = \frac{2 π}{2}$

Divide $x$ by $1$ .

$x = \frac{2 π}{2}$

Simplify the right side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$x = \frac{2 π}{2}$

Divide $π$ by $1$ .

$x = π$

The basic period for $y = 3 \csc (2 x)$ will occur at $(0, π)$ , where $0$ and $π$ are vertical asymptotes.

$(0, π)$

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

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The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 π}{2}$

Divide $π$ by $1$ .

$π$

The vertical asymptotes for $y = 3 \csc (2 x)$ occur at $0$ , $π$ , and every $x = \frac{π n}{2}$ , where $n$ is an integer. This is half of the period.

$x = \frac{π n}{2}$

Cosecant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π n}{2}$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π n}{2}$ where $n$ is an integer

Step 2

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

$a = 3$

$b = 2$

$c = 0$

$d = 0$

Step 3

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

Amplitude: None

Step 4

Find the period of $3 \csc (2 x)$ .

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The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 π}{2}$

Divide $π$ by $1$ .

$π$

Step 5

Find the phase shift using the formula $\frac{c}{b}$ .

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The phase shift of the function can be calculated from $\frac{c}{b}$ .

Phase Shift: $\frac{c}{b}$

Replace the values of $c$ and $b$ in the equation for phase shift.

Phase Shift: $\frac{0}{2}$

Divide $0$ by $2$ .

Phase Shift: $0$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $π$

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: $x = \frac{π n}{2}$ where $n$ is an integer

Amplitude: None

Period: $π$

Phase Shift: None

Vertical Shift: None

Step 8

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