Trigonometry Examples

$y = 3 \cot (2 x + \frac{π}{2})$

Step 1

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 $y = 3 \cot (2 x + \frac{π}{2})$ . 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 $y = 3 \cot (2 x + \frac{π}{2})$ .

$2 x + \frac{π}{2} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $\frac{π}{2}$ from both sides of the equation.

$2 x = - \frac{π}{2}$

Step 2.2

Divide each term in $2 x = - \frac{π}{2}$ by $2$ and simplify.

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Step 2.2.1

Divide each term in $2 x = - \frac{π}{2}$ by $2$ .

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

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.2.1.1

Cancel the common factor.

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

Step 2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \frac{π}{2}}{2}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = - \frac{π}{2} \cdot \frac{1}{2}$

Step 2.2.3.2

Multiply $- \frac{π}{2} \cdot \frac{1}{2}$ .

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Step 2.2.3.2.1

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

$x = - \frac{π}{2 \cdot 2}$

Step 2.2.3.2.2

Multiply $2$ by $2$ .

$x = - \frac{π}{4}$

Step 3

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

$2 x + \frac{π}{2} = π$

Step 4

Solve for $x$ .

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Step 4.1

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

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Step 4.1.1

Subtract $\frac{π}{2}$ from both sides of the equation.

$2 x = π - \frac{π}{2}$

Step 4.1.2

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

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

Step 4.1.3

Combine $π$ and $\frac{2}{2}$ .

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

Step 4.1.4

Combine the numerators over the common denominator.

$2 x = \frac{π \cdot 2 - π}{2}$

Step 4.1.5

Simplify the numerator.

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Step 4.1.5.1

Move $2$ to the left of $π$ .

$2 x = \frac{2 \cdot π - π}{2}$

Step 4.1.5.2

Subtract $π$ from $2 π$ .

$2 x = \frac{π}{2}$

Step 4.2

Divide each term in $2 x = \frac{π}{2}$ by $2$ and simplify.

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Step 4.2.1

Divide each term in $2 x = \frac{π}{2}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{π}{2}}{2}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{π}{2}}{2}$

Step 4.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{2}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{2}$

Step 4.2.3.2

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

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Step 4.2.3.2.1

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

$x = \frac{π}{2 \cdot 2}$

Step 4.2.3.2.2

Multiply $2$ by $2$ .

$x = \frac{π}{4}$

Step 5

The basic period for $y = 3 \cot (2 x + \frac{π}{2})$ will occur at $(- \frac{π}{4}, \frac{π}{4})$ , where $- \frac{π}{4}$ and $\frac{π}{4}$ are vertical asymptotes.

$(- \frac{π}{4}, \frac{π}{4})$

Step 6

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{π}{2}$

Step 7

The vertical asymptotes for $y = 3 \cot (2 x + \frac{π}{2})$ occur at $- \frac{π}{4}$ , $\frac{π}{4}$ , and every $x = - \frac{π}{4} + \frac{π n}{2}$ , where $n$ is an integer.

$x = - \frac{π}{4} + \frac{π n}{2}$

Step 8

Cotangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = - \frac{π}{4} + \frac{π n}{2}$ where $n$ is an integer

Step 9