Trigonometry Examples

y = tan (\frac{2 π}{4} x)

$y = \tan (\frac{2 π}{4} x)$

Step 1

Simplify

tan (\frac{2 π}{4} x)

$\tan (\frac{2 π}{4} x)$ .

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Step 1.1

Cancel the common factor of

2

$2$ and

4

$4$ .

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Step 1.1.1

Factor

2

$2$ out of

2 π

$2 π$ .

y = tan (\frac{2 (π)}{4} x)

$y = \tan (\frac{2 (π)}{4} x)$

Step 1.1.2

Cancel the common factors.

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Step 1.1.2.1

Factor

2

$2$ out of

4

$4$ .

y = tan (\frac{2 π}{2 \cdot 2} x)

$y = \tan (\frac{2 π}{2 \cdot 2} x)$

Step 1.1.2.2

Cancel the common factor.

$y = \tan (\frac{2 π}{2 \cdot 2} x)$

Step 1.1.2.3

Rewrite the expression.

$y = \tan (\frac{π}{2} x)$

Step 1.2

Combine

$\frac{π}{2}$ and

$x$ .

$y = \tan (\frac{π x}{2})$

Step 2

For any

$y = \tan (x)$ , vertical asymptotes occur at

$x = \frac{π}{2} + n π$ , where

$n$ is an integer. Use the basic period for

$y = \tan (x)$ ,

$(- \frac{π}{2}, \frac{π}{2})$ , to find the vertical asymptotes for

$y = \tan (\frac{π x}{2})$ . Set the inside of the tangent function,

$b x + c$ , for

$y = a \tan (b x + c) + d$ equal to

$- \frac{π}{2}$ to find where the vertical asymptote occurs for

$y = \tan (\frac{π x}{2})$ .

$\frac{π x}{2} = - \frac{π}{2}$

Step 3

Solve for

$x$ .

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Step 3.1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$π x = - π$

Step 3.2

Divide each term in

$π x = - π$ by

$π$ and simplify.

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Step 3.2.1

Divide each term in

$π x = - π$ by

$π$ .

$\frac{π x}{π} = \frac{- π}{π}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of

$π$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{π x}{π} = \frac{- π}{π}$

Step 3.2.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{- π}{π}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Cancel the common factor of

$π$ .

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Step 3.2.3.1.1

Cancel the common factor.

$x = \frac{- π}{π}$

Step 3.2.3.1.2

Divide

$- 1$ by

$1$ .

$x = - 1$

Step 4

Set the inside of the tangent function

$\frac{π x}{2}$ equal to

$\frac{π}{2}$ .

$\frac{π x}{2} = \frac{π}{2}$

Step 5

Solve for

$x$ .

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Step 5.1

Since the expression on each side of the equation has the same denominator, the numerators must be equal.

$π x = π$

Step 5.2

Divide each term in

$π x = π$ by

$π$ and simplify.

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Step 5.2.1

Divide each term in

$π x = π$ by

$π$ .

$\frac{π x}{π} = \frac{π}{π}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of

$π$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{π x}{π} = \frac{π}{π}$

Step 5.2.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{π}{π}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Cancel the common factor of

$π$ .

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Step 5.2.3.1.1

Cancel the common factor.

$x = \frac{π}{π}$

Step 5.2.3.1.2

Rewrite the expression.

$x = 1$

Step 6

The basic period for

$y = \tan (\frac{π x}{2})$ will occur at

$(- 1, 1)$ , where

$- 1$ and

$1$ are vertical asymptotes.

$(- 1, 1)$

Step 7

Find the period

$\frac{π}{| b |}$ to find where the vertical asymptotes exist.

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Step 7.1

$\frac{π}{2}$ is approximately

$1.57079632$ which is positive so remove the absolute value

$\frac{π}{\frac{π}{2}}$

Step 7.2

Multiply the numerator by the reciprocal of the denominator.

$π \frac{2}{π}$

Step 7.3

Cancel the common factor of

$π$ .

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Step 7.3.1

Cancel the common factor.

$π \frac{2}{π}$

Step 7.3.2

Rewrite the expression.

$2$

Step 8

The vertical asymptotes for

$y = \tan (\frac{π x}{2})$ occur at

$- 1$ ,

$1$ , and every

$2 n$ , where

$n$ is an integer.

$x = 1 + 2 n$

Step 9

Tangent only has vertical asymptotes.

No Horizontal Asymptotes

No Oblique Asymptotes

Vertical Asymptotes:

$x = 1 + 2 n$ where

$n$ is an integer

Step 10