Trigonometry Examples

$y = \sec (\frac{1}{2} x - 2 π) - 3$

Step 1

Find the asymptotes.

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Step 1.1

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

$\frac{x}{2} - 2 π = - \frac{π}{2}$

Step 1.2

Solve for $x$ .

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Step 1.2.1

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

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Step 1.2.1.1

Add $2 π$ to both sides of the equation.

$\frac{x}{2} = - \frac{π}{2} + 2 π$

Step 1.2.1.2

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

$\frac{x}{2} = - \frac{π}{2} + 2 π \cdot \frac{2}{2}$

Step 1.2.1.3

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{x}{2} = - \frac{π}{2} + \frac{2 π \cdot 2}{2}$

Step 1.2.1.4

Combine the numerators over the common denominator.

$\frac{x}{2} = \frac{- π + 2 π \cdot 2}{2}$

Step 1.2.1.5

Simplify the numerator.

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Step 1.2.1.5.1

Multiply $2$ by $2$ .

$\frac{x}{2} = \frac{- π + 4 π}{2}$

Step 1.2.1.5.2

Add $- π$ and $4 π$ .

$\frac{x}{2} = \frac{3 π}{2}$

Step 1.2.2

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = 3 π$

Step 1.3

Set the inside of the secant function $\frac{x}{2} - 2 π$ equal to $\frac{3 π}{2}$ .

$\frac{x}{2} - 2 π = \frac{3 π}{2}$

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 $2 π$ to both sides of the equation.

$\frac{x}{2} = \frac{3 π}{2} + 2 π$

Step 1.4.1.2

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

$\frac{x}{2} = \frac{3 π}{2} + 2 π \cdot \frac{2}{2}$

Step 1.4.1.3

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{x}{2} = \frac{3 π}{2} + \frac{2 π \cdot 2}{2}$

Step 1.4.1.4

Combine the numerators over the common denominator.

$\frac{x}{2} = \frac{3 π + 2 π \cdot 2}{2}$

Step 1.4.1.5

Simplify the numerator.

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Step 1.4.1.5.1

Multiply $2$ by $2$ .

$\frac{x}{2} = \frac{3 π + 4 π}{2}$

Step 1.4.1.5.2

Add $3 π$ and $4 π$ .

$\frac{x}{2} = \frac{7 π}{2}$

Step 1.4.2

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$x = 7 π$

Step 1.5

The basic period for $y = \sec (\frac{x}{2} - 2 π) - 3$ will occur at $(3 π, 7 π)$ , where $3 π$ and $7 π$ are vertical asymptotes.

$(3 π, 7 π)$

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

$\frac{1}{2}$ is approximately $0.5$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{2}}$

Step 1.6.2

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 2$

Step 1.6.3

Multiply $2$ by $2$ .

$4 π$

Step 1.7

The vertical asymptotes for $y = \sec (\frac{x}{2} - 2 π) - 3$ occur at $3 π$ , $7 π$ , and every $x = 3 π + 2 π n$ , where $n$ is an integer. This is half of the period.

$x = 3 π + 2 π n$

Step 1.8

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = 3 π + 2 π n$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = 3 π + 2 π n$ where $n$ is an integer

Step 2

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

$a = 1$

$b = \frac{1}{2}$

$c = 2 π$

$d = - 3$

Step 3

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

Amplitude: None

Step 4

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

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Step 4.1

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

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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 $\frac{1}{2}$ in the formula for period.

$\frac{2 π}{| \frac{1}{2} |}$

Step 4.1.3

$\frac{1}{2}$ is approximately $0.5$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{2}}$

Step 4.1.4

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 2$

Step 4.1.5

Multiply $2$ by $2$ .

$4 π$

Step 4.2

Find the period of $- 3$ .

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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 $\frac{1}{2}$ in the formula for period.

$\frac{2 π}{| \frac{1}{2} |}$

Step 4.2.3

$\frac{1}{2}$ is approximately $0.5$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{2}}$

Step 4.2.4

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 2$

Step 4.2.5

Multiply $2$ by $2$ .

$4 π$

Step 4.3

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

$4 π$

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: $\frac{2 π}{\frac{1}{2}}$

Step 5.3

Multiply the numerator by the reciprocal of the denominator.

Phase Shift: $2 π \cdot 2$

Step 5.4

Multiply $2$ by $2$ .

Phase Shift: $4 π$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $4 π$

Phase Shift: $4 π$ ( $4 π$ to the right)

Vertical Shift: $- 3$

Step 7

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

Vertical Asymptotes: $x = 3 π + 2 π n$ where $n$ is an integer

Amplitude: None

Period: $4 π$

Phase Shift: $4 π$ ( $4 π$ to the right)

Vertical Shift: $- 3$

Step 8