Trigonometry Examples

$f (t) = 2 \csc (4 t - \frac{π}{5}) + 1$

Step 1

Find the asymptotes.

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Step 1.1

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 $y = 2 \csc (4 x - \frac{π}{5}) + 1$ . 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 $y = 2 \csc (4 x - \frac{π}{5}) + 1$ .

$4 x - \frac{π}{5} = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Add $\frac{π}{5}$ to both sides of the equation.

$4 x = \frac{π}{5}$

Step 1.2.2

Divide each term in $4 x = \frac{π}{5}$ by $4$ and simplify.

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Step 1.2.2.1

Divide each term in $4 x = \frac{π}{5}$ by $4$ .

$\frac{4 x}{4} = \frac{\frac{π}{5}}{4}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Cancel the common factor of $4$ .

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Step 1.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{\frac{π}{5}}{4}$

Step 1.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{5}}{4}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{5} \cdot \frac{1}{4}$

Step 1.2.2.3.2

Multiply $\frac{π}{5} \cdot \frac{1}{4}$ .

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Step 1.2.2.3.2.1

Multiply $\frac{π}{5}$ by $\frac{1}{4}$ .

$x = \frac{π}{5 \cdot 4}$

Step 1.2.2.3.2.2

Multiply $5$ by $4$ .

$x = \frac{π}{20}$

Step 1.3

Set the inside of the cosecant function $4 x - \frac{π}{5}$ equal to $2 π$ .

$4 x - \frac{π}{5} = 2 π$

Step 1.4

Solve for $x$ .

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Step 1.4.1

Move all terms not containing $x$ to the right side of the equation.

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Step 1.4.1.1

Add $\frac{π}{5}$ to both sides of the equation.

$4 x = 2 π + \frac{π}{5}$

Step 1.4.1.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$4 x = 2 π \cdot \frac{5}{5} + \frac{π}{5}$

Step 1.4.1.3

Combine $2 π$ and $\frac{5}{5}$ .

$4 x = \frac{2 π \cdot 5}{5} + \frac{π}{5}$

Step 1.4.1.4

Combine the numerators over the common denominator.

$4 x = \frac{2 π \cdot 5 + π}{5}$

Step 1.4.1.5

Simplify the numerator.

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Step 1.4.1.5.1

Multiply $5$ by $2$ .

$4 x = \frac{10 π + π}{5}$

Step 1.4.1.5.2

Add $10 π$ and $π$ .

$4 x = \frac{11 π}{5}$

Step 1.4.2

Divide each term in $4 x = \frac{11 π}{5}$ by $4$ and simplify.

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Step 1.4.2.1

Divide each term in $4 x = \frac{11 π}{5}$ by $4$ .

$\frac{4 x}{4} = \frac{\frac{11 π}{5}}{4}$

Step 1.4.2.2

Simplify the left side.

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Step 1.4.2.2.1

Cancel the common factor of $4$ .

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Step 1.4.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{\frac{11 π}{5}}{4}$

Step 1.4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{11 π}{5}}{4}$

Step 1.4.2.3

Simplify the right side.

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Step 1.4.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{11 π}{5} \cdot \frac{1}{4}$

Step 1.4.2.3.2

Multiply $\frac{11 π}{5} \cdot \frac{1}{4}$ .

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Step 1.4.2.3.2.1

Multiply $\frac{11 π}{5}$ by $\frac{1}{4}$ .

$x = \frac{11 π}{5 \cdot 4}$

Step 1.4.2.3.2.2

Multiply $5$ by $4$ .

$x = \frac{11 π}{20}$

Step 1.5

The basic period for $y = 2 \csc (4 x - \frac{π}{5}) + 1$ will occur at $(\frac{π}{20}, \frac{11 π}{20})$ , where $\frac{π}{20}$ and $\frac{11 π}{20}$ are vertical asymptotes.

$(\frac{π}{20}, \frac{11 π}{20})$

Step 1.6

Find the period $\frac{2 π}{| b |}$ 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 $0$ and $4$ is $4$ .

$\frac{2 π}{4}$

Step 1.6.2

Cancel the common factor of $2$ and $4$ .

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Step 1.6.2.1

Factor $2$ out of $2 π$ .

$\frac{2 (π)}{4}$

Step 1.6.2.2

Cancel the common factors.

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Step 1.6.2.2.1

Factor $2$ out of $4$ .

$\frac{2 π}{2 \cdot 2}$

Step 1.6.2.2.2

Cancel the common factor.

$\frac{2 π}{2 \cdot 2}$

Step 1.6.2.2.3

Rewrite the expression.

$\frac{π}{2}$

Step 1.7

The vertical asymptotes for $y = 2 \csc (4 x - \frac{π}{5}) + 1$ occur at $\frac{π}{20}$ , $\frac{11 π}{20}$ , and every $x = \frac{π}{20} + \frac{π n}{4}$ , where $n$ is an integer. This is half of the period.

$x = \frac{π}{20} + \frac{π n}{4}$

Step 1.8

Cosecant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π}{20} + \frac{π n}{4}$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π}{20} + \frac{π n}{4}$ 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.

$a = 2$

$b = 4$

$c = \frac{π}{5}$

$d = 1$

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 using the formula $\frac{2 π}{| b |}$ .

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Step 4.1

Find the period of $2 \csc (4 x - \frac{π}{5})$ .

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Step 4.1.1

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

$\frac{2 π}{| b |}$

Step 4.1.2

Replace $b$ with $4$ in the formula for period.

$\frac{2 π}{| 4 |}$

Step 4.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$\frac{2 π}{4}$

Step 4.1.4

Cancel the common factor of $2$ and $4$ .

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Step 4.1.4.1

Factor $2$ out of $2 π$ .

$\frac{2 (π)}{4}$

Step 4.1.4.2

Cancel the common factors.

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Step 4.1.4.2.1

Factor $2$ out of $4$ .

$\frac{2 π}{2 \cdot 2}$

Step 4.1.4.2.2

Cancel the common factor.

$\frac{2 π}{2 \cdot 2}$

Step 4.1.4.2.3

Rewrite the expression.

$\frac{π}{2}$

Step 4.2

Find the period of $1$ .

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Step 4.2.1

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

$\frac{2 π}{| b |}$

Step 4.2.2

Replace $b$ with $4$ in the formula for period.

$\frac{2 π}{| 4 |}$

Step 4.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $4$ is $4$ .

$\frac{2 π}{4}$

Step 4.2.4

Cancel the common factor of $2$ and $4$ .

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Step 4.2.4.1

Factor $2$ out of $2 π$ .

$\frac{2 (π)}{4}$

Step 4.2.4.2

Cancel the common factors.

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Step 4.2.4.2.1

Factor $2$ out of $4$ .

$\frac{2 π}{2 \cdot 2}$

Step 4.2.4.2.2

Cancel the common factor.

$\frac{2 π}{2 \cdot 2}$

Step 4.2.4.2.3

Rewrite the expression.

$\frac{π}{2}$

Step 4.3

The period of addition/subtraction of trig functions is the maximum of the individual periods.

$\frac{π}{2}$

Step 5

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

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Step 5.1

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

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

Step 5.2

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

Phase Shift: $\frac{\frac{π}{5}}{4}$

Step 5.3

Multiply the numerator by the reciprocal of the denominator.

Phase Shift: $\frac{π}{5} \cdot \frac{1}{4}$

Step 5.4

Multiply $\frac{π}{5} \cdot \frac{1}{4}$ .

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Step 5.4.1

Multiply $\frac{π}{5}$ by $\frac{1}{4}$ .

Phase Shift: $\frac{π}{5 \cdot 4}$

Step 5.4.2

Multiply $5$ by $4$ .

Phase Shift: $\frac{π}{20}$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{π}{2}$

Phase Shift: $\frac{π}{20}$ ( $\frac{π}{20}$ to the right)

Vertical Shift: $1$

Step 7

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

Vertical Asymptotes: $x = \frac{π}{20} + \frac{π n}{4}$ where $n$ is an integer

Amplitude: None

Period: $\frac{π}{2}$

Phase Shift: $\frac{π}{20}$ ( $\frac{π}{20}$ to the right)

Vertical Shift: $1$

Step 8