Trigonometry Examples

$f (x) = 2 \cot (4 x)$

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 $y = 2 \cot (4 x)$ . Set the inside of the cotangent function, bx+c, for $y = a \cot (b x + c) + d$ equal to $0$ to find where the vertical asymptote occurs for $y = 2 \cot (4 x)$ .

$4 x = 0$

Divide each term by $4$ and simplify.

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Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Divide $0$ by $4$ .

$x = 0$

Set the inside of the cotangent function $4 x$ equal to $π$ .

$4 x = π$

Divide each term by $4$ and simplify.

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Divide each term in $4 x = π$ by $4$ .

$\frac{4 x}{4} = \frac{π}{4}$

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

$\frac{4 x}{4} = \frac{π}{4}$

Divide $x$ by $1$ .

$x = \frac{π}{4}$

The basic period for $y = 2 \cot (4 x)$ will occur at $(0, \frac{π}{4})$ , where $0$ and $\frac{π}{4}$ are vertical asymptotes.

$(0, \frac{π}{4})$

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

$\frac{π}{4}$

The vertical asymptotes for $y = 2 \cot (4 x)$ occur at $0$ , $\frac{π}{4}$ , and every $x = \frac{π n}{4}$ , where $n$ is an integer.

$x = \frac{π n}{4}$

Cotangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

No Horizontal Asymptotes

No Oblique Asymptotes

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

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

$a = 2$

$b = 4$

$c = 0$

$d = 0$

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 $4$ in the formula for period.

Period: $\frac{π}{| 4 |}$

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

Period: $\frac{π}{4}$

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: $\frac{0}{4}$

Divide $0$ by $4$ .

Phase Shift: $0$

Find the vertical shift $d$ .

Vertical Shift: $0$

List the properties of the trigonometric function.

Amplitude: None

Period: $\frac{π}{4}$

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

Vertical Shift: $0$

Select a few points to graph.

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Find the point at $x = \frac{π}{16}$ .

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Replace the variable $x$ with $\frac{π}{16}$ in the expression.

$f (\frac{π}{16}) = 2 \cot (4 (\frac{π}{16}))$

Simplify the result.

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Cancel the common factor of $4$ .

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Write $4$ as a fraction with denominator $1$ .

$f (\frac{π}{16}) = 2 \cot (\frac{4}{1} \cdot \frac{π}{16})$

Factor out the greatest common factor $4$ .

$f (\frac{π}{16}) = 2 \cot (\frac{4 \cdot 1}{1} \cdot \frac{π}{4 \cdot 4})$

Cancel the common factor.

$f (\frac{π}{16}) = 2 \cot (\frac{4 \cdot 1}{1} \cdot \frac{π}{4 \cdot 4})$

Rewrite the expression.

$f (\frac{π}{16}) = 2 \cot (\frac{1}{1} \cdot \frac{π}{4})$

Multiply $\frac{1}{1}$ and $\frac{π}{4}$ .

$f (\frac{π}{16}) = 2 \cot (\frac{π}{4})$

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

$f (\frac{π}{16}) = 2 \cdot 1$

Multiply $2$ by $1$ .

$f (\frac{π}{16}) = 2$

The final answer is $2$ .

$2$

Find the point at $x = \frac{π}{8}$ .

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Replace the variable $x$ with $\frac{π}{8}$ in the expression.

$f (\frac{π}{8}) = 2 \cot (4 (\frac{π}{8}))$

Simplify the result.

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Cancel the common factor of $4$ .

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Write $4$ as a fraction with denominator $1$ .

$f (\frac{π}{8}) = 2 \cot (\frac{4}{1} \cdot \frac{π}{8})$

Factor out the greatest common factor $4$ .

$f (\frac{π}{8}) = 2 \cot (\frac{4 \cdot 1}{1} \cdot \frac{π}{4 \cdot 2})$

Cancel the common factor.

$f (\frac{π}{8}) = 2 \cot (\frac{4 \cdot 1}{1} \cdot \frac{π}{4 \cdot 2})$

Rewrite the expression.

$f (\frac{π}{8}) = 2 \cot (\frac{1}{1} \cdot \frac{π}{2})$

Multiply $\frac{1}{1}$ and $\frac{π}{2}$ .

$f (\frac{π}{8}) = 2 \cot (\frac{π}{2})$

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

$f (\frac{π}{8}) = 2 \cdot 0$

Multiply $2$ by $0$ .

$f (\frac{π}{8}) = 0$

The final answer is $0$ .

$0$

Find the point at $x = \frac{3 π}{16}$ .

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Replace the variable $x$ with $\frac{3 π}{16}$ in the expression.

$f (\frac{3 π}{16}) = 2 \cot (4 (\frac{3 π}{16}))$

Simplify the result.

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Cancel the common factor of $4$ .

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Write $4$ as a fraction with denominator $1$ .

$f (\frac{3 π}{16}) = 2 \cot (\frac{4}{1} \cdot \frac{3 π}{16})$

Factor out the greatest common factor $4$ .

$f (\frac{3 π}{16}) = 2 \cot (\frac{4 \cdot 1}{1} \cdot \frac{3 π}{4 \cdot 4})$

Cancel the common factor.

$f (\frac{3 π}{16}) = 2 \cot (\frac{4 \cdot 1}{1} \cdot \frac{3 π}{4 \cdot 4})$

Rewrite the expression.

$f (\frac{3 π}{16}) = 2 \cot (\frac{1}{1} \cdot \frac{3 π}{4})$

Multiply $\frac{1}{1}$ and $\frac{3 π}{4}$ .

$f (\frac{3 π}{16}) = 2 \cot (\frac{3 π}{4})$

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.

$f (\frac{3 π}{16}) = 2 (- \cot (\frac{π}{4}))$

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

$f (\frac{3 π}{16}) = 2 (- 1 \cdot 1)$

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{16}) = 2 \cdot - 1$

Multiply $2$ by $- 1$ .

$f (\frac{3 π}{16}) = - 2$

The final answer is $- 2$ .

$- 2$

Find the point at $x = \frac{5 π}{16}$ .

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Replace the variable $x$ with $\frac{5 π}{16}$ in the expression.

$f (\frac{5 π}{16}) = 2 \cot (4 (\frac{5 π}{16}))$

Simplify the result.

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Cancel the common factor of $4$ .

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Write $4$ as a fraction with denominator $1$ .

$f (\frac{5 π}{16}) = 2 \cot (\frac{4}{1} \cdot \frac{5 π}{16})$

Factor out the greatest common factor $4$ .

$f (\frac{5 π}{16}) = 2 \cot (\frac{4 \cdot 1}{1} \cdot \frac{5 π}{4 \cdot 4})$

Cancel the common factor.

$f (\frac{5 π}{16}) = 2 \cot (\frac{4 \cdot 1}{1} \cdot \frac{5 π}{4 \cdot 4})$

Rewrite the expression.

$f (\frac{5 π}{16}) = 2 \cot (\frac{1}{1} \cdot \frac{5 π}{4})$

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

$f (\frac{5 π}{16}) = 2 \cot (\frac{5 π}{4})$

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

$f (\frac{5 π}{16}) = 2 \cot (\frac{π}{4})$

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

$f (\frac{5 π}{16}) = 2 \cdot 1$

Multiply $2$ by $1$ .

$f (\frac{5 π}{16}) = 2$

The final answer is $2$ .

$2$

Find the point at $x = \frac{3 π}{8}$ .

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Replace the variable $x$ with $\frac{3 π}{8}$ in the expression.

$f (\frac{3 π}{8}) = 2 \cot (4 (\frac{3 π}{8}))$

Simplify the result.

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Cancel the common factor of $4$ .

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Write $4$ as a fraction with denominator $1$ .

$f (\frac{3 π}{8}) = 2 \cot (\frac{4}{1} \cdot \frac{3 π}{8})$

Factor out the greatest common factor $4$ .

$f (\frac{3 π}{8}) = 2 \cot (\frac{4 \cdot 1}{1} \cdot \frac{3 π}{4 \cdot 2})$

Cancel the common factor.

$f (\frac{3 π}{8}) = 2 \cot (\frac{4 \cdot 1}{1} \cdot \frac{3 π}{4 \cdot 2})$

Rewrite the expression.

$f (\frac{3 π}{8}) = 2 \cot (\frac{1}{1} \cdot \frac{3 π}{2})$

Multiply $\frac{1}{1}$ and $\frac{3 π}{2}$ .

$f (\frac{3 π}{8}) = 2 \cot (\frac{3 π}{2})$

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.

$f (\frac{3 π}{8}) = 2 (- \cot (\frac{π}{2}))$

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

$f (\frac{3 π}{8}) = 2 (- 0)$

Multiply $- 1$ by $0$ .

$f (\frac{3 π}{8}) = 2 \cdot 0$

Multiply $2$ by $0$ .

$f (\frac{3 π}{8}) = 0$

The final answer is $0$ .

$0$

List the points in a table.

$\begin{array}{c} x & f (x) \\ \frac{π}{16} & 2 \\ \frac{π}{8} & 0 \\ \frac{3 π}{16} & - 2 \\ \frac{5 π}{16} & 2 \\ \frac{3 π}{8} & 0 \end{array}$

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

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

Amplitude: None

Period: $\frac{π}{4}$

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

Vertical Shift: $0$

$\begin{array}{c} x & f (x) \\ \frac{π}{16} & 2 \\ \frac{π}{8} & 0 \\ \frac{3 π}{16} & - 2 \\ \frac{5 π}{16} & 2 \\ \frac{3 π}{8} & 0 \end{array}$

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