Trigonometry Examples

$\sec (\frac{π}{2} - x)$

Step 1

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 $y = \sec (\frac{π}{2} - x)$ . 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 $y = \sec (\frac{π}{2} - x)$ .

$\frac{π}{2} - x = - \frac{π}{2}$

Step 2

Solve for $x$ .

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Step 2.1

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

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Step 2.1.1

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

$- x = - \frac{π}{2} - \frac{π}{2}$

Step 2.1.2

Combine the numerators over the common denominator.

$- x = \frac{- π - π}{2}$

Step 2.1.3

Subtract $π$ from $- π$ .

$- x = \frac{- 2 π}{2}$

Step 2.1.4

Cancel the common factor of $- 2$ and $2$ .

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Step 2.1.4.1

Factor $2$ out of $- 2 π$ .

$- x = \frac{2 (- π)}{2}$

Step 2.1.4.2

Cancel the common factors.

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Step 2.1.4.2.1

Factor $2$ out of $2$ .

$- x = \frac{2 (- π)}{2 (1)}$

Step 2.1.4.2.2

Cancel the common factor.

$- x = \frac{2 (- π)}{2 \cdot 1}$

Step 2.1.4.2.3

Rewrite the expression.

$- x = \frac{- π}{1}$

Step 2.1.4.2.4

Divide $- π$ by $1$ .

$- x = - π$

Step 2.2

Divide each term in $- x = - π$ by $- 1$ and simplify.

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Step 2.2.1

Divide each term in $- x = - π$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- π}{- 1}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- π}{- 1}$

Step 2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- π}{- 1}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Dividing two negative values results in a positive value.

$x = \frac{π}{1}$

Step 2.2.3.2

Divide $π$ by $1$ .

$x = π$

Step 3

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

$\frac{π}{2} - x = \frac{3 π}{2}$

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.

$- x = \frac{3 π}{2} - \frac{π}{2}$

Step 4.1.2

Combine the numerators over the common denominator.

$- x = \frac{3 π - π}{2}$

Step 4.1.3

Subtract $π$ from $3 π$ .

$- x = \frac{2 π}{2}$

Step 4.1.4

Cancel the common factor of $2$ .

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Step 4.1.4.1

Cancel the common factor.

$- x = \frac{2 π}{2}$

Step 4.1.4.2

Divide $π$ by $1$ .

$- x = π$

Step 4.2

Divide each term in $- x = π$ by $- 1$ and simplify.

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Step 4.2.1

Divide each term in $- x = π$ by $- 1$ .

$\frac{- x}{- 1} = \frac{π}{- 1}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{π}{- 1}$

Step 4.2.2.2

Divide $x$ by $1$ .

$x = \frac{π}{- 1}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

Move the negative one from the denominator of $\frac{π}{- 1}$ .

$x = - 1 \cdot π$

Step 4.2.3.2

Rewrite $- 1 \cdot π$ as $- π$ .

$x = - π$

Step 5

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

$(π, - π)$

Step 6

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 6.1

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

$\frac{2 π}{1}$

Step 6.2

Divide $2 π$ by $1$ .

$2 π$

Step 7

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

$x = π + π n$

Step 8

Secant only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes: $x = π + π n$ where $n$ is an integer

Step 9