Trigonometry Examples

$r \cos (x) = - 6$

Step 1

Find the asymptotes.

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Step 1.1

Divide each term in $y \cos (x) = - 6$ by $\cos (x)$ and simplify.

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Step 1.1.1

Divide each term in $y \cos (x) = - 6$ by $\cos (x)$ .

$\frac{y \cos (x)}{\cos (x)} = \frac{- 6}{\cos (x)}$

Step 1.1.2

Simplify the left side.

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Step 1.1.2.1

Cancel the common factor of $\cos (x)$ .

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Step 1.1.2.1.1

Cancel the common factor.

$\frac{y \cos (x)}{\cos (x)} = \frac{- 6}{\cos (x)}$

Step 1.1.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 6}{\cos (x)}$

Step 1.1.3

Simplify the right side.

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Step 1.1.3.1

Separate fractions.

$y = \frac{- 6}{1} \cdot \frac{1}{\cos (x)}$

Step 1.1.3.2

Convert from $\frac{1}{\cos (x)}$ to $\sec (x)$ .

$y = \frac{- 6}{1} \sec (x)$

Step 1.1.3.3

Divide $- 6$ by $1$ .

$y = - 6 \sec (x)$

Step 1.2

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 $y = - 6 \sec (x)$ . 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 $y = - 6 \sec (x)$ .

$x = - \frac{π}{2}$

Step 1.3

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

$x = \frac{3 π}{2}$

Step 1.4

The basic period for $y = - 6 \sec (x)$ will occur at $(- \frac{π}{2}, \frac{3 π}{2})$ , where $- \frac{π}{2}$ and $\frac{3 π}{2}$ are vertical asymptotes.

$(- \frac{π}{2}, \frac{3 π}{2})$

Step 1.5

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 1.5.1

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 1.5.2

Divide $2 π$ by $1$ .

$2 π$

Step 1.6

The vertical asymptotes for $y = - 6 \sec (x)$ occur at $- \frac{π}{2}$ , $\frac{3 π}{2}$ , and every $π n$ , where $n$ is an integer. This is half of the period.

$π n$

Step 1.7

There are only vertical asymptotes for secant and cosecant functions.

Vertical Asymptotes: $x = \frac{3 π}{2} + π n$ for any integer $n$

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{3 π}{2} + π n$ for any integer $n$

No Horizontal Asymptotes

No Oblique Asymptotes

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.

$a = - 6$

$b = 1$

$c = 0$

$d = 0$

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 of $- 6 \sec (x)$ .

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Step 4.1

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

$\frac{2 π}{| b |}$

Step 4.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 4.4

Divide $2 π$ by $1$ .

$2 π$

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: $\frac{0}{1}$

Step 5.3

Divide $0$ by $1$ .

Phase Shift: $0$

Step 6

List the properties of the trigonometric function.

Amplitude: None

Period: $2 π$

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: $x = \frac{3 π}{2} + π n$ for any integer $n$

Amplitude: None

Period: $2 π$

Phase Shift: None

Vertical Shift: None

Step 8