Trigonometry Examples

$f (x) = 4 \cot (3 x)$

Find the asymptotes.

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For any $y = \cot (x)$ , vertical asymptotes occur at $x = n π$ , where $n$ is an integer. Use the basic period for $y = \cot (x)$ , $(0, π)$ , to find the vertical asymptotes for $y = 4 \cot (3 x)$ . Set the inside of the cotangent function, $b x + c$ , for $y = a \cot (b x + c) + d$ equal to $0$ to find where the vertical asymptote occurs for $y = 4 \cot (3 x)$ .

$3 x = 0$

Divide each term by $3$ and simplify.

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Divide each term in $3 x = 0$ by $3$ .

$\frac{3 x}{3} = \frac{0}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 x}{3} = \frac{0}{3}$

Divide $x$ by $1$ .

$x = \frac{0}{3}$

Divide $0$ by $3$ .

$x = 0$

Set the inside of the cotangent function $3 x$ equal to $π$ .

$3 x = π$

Divide each term by $3$ and simplify.

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Divide each term in $3 x = π$ by $3$ .

$\frac{3 x}{3} = \frac{π}{3}$

Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 x}{3} = \frac{π}{3}$

Divide $x$ by $1$ .

$x = \frac{π}{3}$

The basic period for $y = 4 \cot (3 x)$ will occur at $(0, \frac{π}{3})$ , where $0$ and $\frac{π}{3}$ are vertical asymptotes.

$(0, \frac{π}{3})$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{π}{3}$

The vertical asymptotes for $y = 4 \cot (3 x)$ occur at $0$ , $\frac{π}{3}$ , and every $\frac{π n}{3}$ , where $n$ is an integer.

$x = \frac{π n}{3}$

Cotangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π n}{3}$ where $n$ is an integer

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = \frac{π n}{3}$ where $n$ is an integer

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

$a = 4$

$b = 3$

$c = 0$

$d = 0$

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

Amplitude: None

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

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

Period: $\frac{π}{| b |}$

Replace $b$ with $3$ in the formula for period.

Period: $\frac{π}{| 3 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

Period: $\frac{π}{3}$

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: $\frac{0}{3}$

Divide $0$ by $3$ .

Phase Shift: $0$

Find the vertical shift $d$ .

Vertical Shift: $0$

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{π}{3}$

Phase Shift: $0$ ( $0$ to the right)

Vertical Shift: $0$

Select a few points to graph.

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Find the point at $x = \frac{π}{12}$ .

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Replace the variable $x$ with $\frac{π}{12}$ in the expression.

$f (\frac{π}{12}) = 4 \cot (3 (\frac{π}{12}))$

Simplify the result.

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Cancel the common factor of $3$ .

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Factor $3$ out of $12$ .

$f (\frac{π}{12}) = 4 \cot (3 (\frac{π}{3 (4)}))$

Cancel the common factor.

$f (\frac{π}{12}) = 4 \cot (3 (\frac{π}{3 \cdot 4}))$

Rewrite the expression.

$f (\frac{π}{12}) = 4 \cot (\frac{π}{4})$

The exact value of $\cot (\frac{π}{4})$ is $1$ .

$f (\frac{π}{12}) = 4 \cdot 1$

Multiply $4$ by $1$ .

$f (\frac{π}{12}) = 4$

The final answer is $4$ .

$4$

Find the point at $x = \frac{π}{6}$ .

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Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = 4 \cot (3 (\frac{π}{6}))$

Simplify the result.

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Cancel the common factor of $3$ .

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Factor $3$ out of $6$ .

$f (\frac{π}{6}) = 4 \cot (3 (\frac{π}{3 (2)}))$

Cancel the common factor.

$f (\frac{π}{6}) = 4 \cot (3 (\frac{π}{3 \cdot 2}))$

Rewrite the expression.

$f (\frac{π}{6}) = 4 \cot (\frac{π}{2})$

The exact value of $\cot (\frac{π}{2})$ is $0$ .

$f (\frac{π}{6}) = 4 \cdot 0$

Multiply $4$ by $0$ .

$f (\frac{π}{6}) = 0$

The final answer is $0$ .

$0$

Find the point at $x = \frac{π}{4}$ .

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Replace the variable $x$ with $\frac{π}{4}$ in the expression.

$f (\frac{π}{4}) = 4 \cot (3 (\frac{π}{4}))$

Simplify the result.

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Combine $3$ and $\frac{π}{4}$ .

$f (\frac{π}{4}) = 4 \cot (\frac{3 π}{4})$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cotangent is negative in the second quadrant.

$f (\frac{π}{4}) = 4 (- \cot (\frac{π}{4}))$

The exact value of $\cot (\frac{π}{4})$ is $1$ .

$f (\frac{π}{4}) = 4 (- 1 \cdot 1)$

Multiply $- 1$ by $1$ .

$f (\frac{π}{4}) = 4 \cdot - 1$

Multiply $4$ by $- 1$ .

$f (\frac{π}{4}) = - 4$

The final answer is $- 4$ .

$- 4$

Find the point at $x = \frac{5 π}{12}$ .

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Replace the variable $x$ with $\frac{5 π}{12}$ in the expression.

$f (\frac{5 π}{12}) = 4 \cot (3 (\frac{5 π}{12}))$

Simplify the result.

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Cancel the common factor of $3$ .

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Factor $3$ out of $12$ .

$f (\frac{5 π}{12}) = 4 \cot (3 (\frac{5 π}{3 (4)}))$

Cancel the common factor.

$f (\frac{5 π}{12}) = 4 \cot (3 (\frac{5 π}{3 \cdot 4}))$

Rewrite the expression.

$f (\frac{5 π}{12}) = 4 \cot (\frac{5 π}{4})$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$f (\frac{5 π}{12}) = 4 \cot (\frac{π}{4})$

The exact value of $\cot (\frac{π}{4})$ is $1$ .

$f (\frac{5 π}{12}) = 4 \cdot 1$

Multiply $4$ by $1$ .

$f (\frac{5 π}{12}) = 4$

The final answer is $4$ .

$4$

Find the point at $x = \frac{π}{2}$ .

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Replace the variable $x$ with $\frac{π}{2}$ in the expression.

$f (\frac{π}{2}) = 4 \cot (3 (\frac{π}{2}))$

Simplify the result.

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Combine $3$ and $\frac{π}{2}$ .

$f (\frac{π}{2}) = 4 \cot (\frac{3 π}{2})$

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cotangent is negative in the fourth quadrant.

$f (\frac{π}{2}) = 4 (- \cot (\frac{π}{2}))$

The exact value of $\cot (\frac{π}{2})$ is $0$ .

$f (\frac{π}{2}) = 4 (- 0)$

Multiply $- 1$ by $0$ .

$f (\frac{π}{2}) = 4 \cdot 0$

Multiply $4$ by $0$ .

$f (\frac{π}{2}) = 0$

The final answer is $0$ .

$0$

List the points in a table.

$\begin{array}{c} x & f (x) \\ \frac{π}{12} & 4 \\ \frac{π}{6} & 0 \\ \frac{π}{4} & - 4 \\ \frac{5 π}{12} & 4 \\ \frac{π}{2} & 0 \end{array}$

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

Vertical Asymptotes: $x = \frac{π n}{3}$ where $n$ is an integer

Amplitude: None

Period: $\frac{π}{3}$

Phase Shift: $0$ ( $0$ to the right)

Vertical Shift: $0$

$\begin{array}{c} x & f (x) \\ \frac{π}{12} & 4 \\ \frac{π}{6} & 0 \\ \frac{π}{4} & - 4 \\ \frac{5 π}{12} & 4 \\ \frac{π}{2} & 0 \end{array}$

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