Trigonometry Examples

$y = \frac{1}{3} \cdot \csc (2 x + π)$

Step 1

Combine $\frac{1}{3}$ and $\csc (2 x + π)$ .

$y = \frac{\csc (2 x + π)}{3}$

Step 2

For any $y = \csc (x)$ , vertical asymptotes occur at $x = n π$ , where $n$ is an integer. Use the basic period for $y = c s c (x)$ , $(0, 2 π)$ , to find the vertical asymptotes for $y = \frac{\csc (2 x + π)}{3}$ . Set the inside of the cosecant function, $b x + c$ , for $y = a \csc (b x + c) + d$ equal to $0$ to find where the vertical asymptote occurs for $y = \frac{\csc (2 x + π)}{3}$ .

$2 x + π = 0$

Step 3

Solve for $x$ .

Tap for more steps...

Step 3.1

Subtract $π$ from both sides of the equation.

$2 x = - π$

Step 3.2

Divide each term in $2 x = - π$ by $2$ and simplify.

Tap for more steps...

Step 3.2.1

Divide each term in $2 x = - π$ by $2$ .

$\frac{2 x}{2} = \frac{- π}{2}$

Step 3.2.2

Simplify the left side.

Tap for more steps...

Step 3.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- π}{2}$

Step 3.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- π}{2}$

Step 3.2.3

Simplify the right side.

Tap for more steps...

Step 3.2.3.1

Move the negative in front of the fraction.

$x = - \frac{π}{2}$

Step 4

Set the inside of the cosecant function $2 x + π$ equal to $2 π$ .

$2 x + π = 2 π$

Step 5

Solve for $x$ .

Tap for more steps...

Step 5.1

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

Tap for more steps...

Step 5.1.1

Subtract $π$ from both sides of the equation.

$2 x = 2 π - π$

Step 5.1.2

Subtract $π$ from $2 π$ .

$2 x = π$

Step 5.2

Divide each term in $2 x = π$ by $2$ and simplify.

Tap for more steps...

Step 5.2.1

Divide each term in $2 x = π$ by $2$ .

$\frac{2 x}{2} = \frac{π}{2}$

Step 5.2.2

Simplify the left side.

Tap for more steps...

Step 5.2.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 5.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Step 5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Step 6

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

$(- \frac{π}{2}, \frac{π}{2})$

Step 7

Find the period $\frac{2 π}{| b |}$ to find where the vertical asymptotes exist. Vertical asymptotes occur every half period.

Tap for more steps...

Step 7.1

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

$\frac{2 π}{2}$

Step 7.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 7.2.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 7.2.2

Divide $π$ by $1$ .

$π$

Step 8

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

$x = - \frac{π}{2} + \frac{π n}{2}$

Step 9

Cosecant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

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

Step 10