Trigonometry Examples

y = - tan (\frac{1}{10} x) + 4

$y = - \tan (\frac{1}{10} x) + 4$

Step 1

The parent function is the simplest form of the type of function given.

y = tan (x)

$y = \tan (x)$

Step 2

Combine

\frac{1}{10}

$\frac{1}{10}$ and

x

$x$ .

y = - tan (\frac{x}{10}) + 4

$y = - \tan (\frac{x}{10}) + 4$

Step 3

Assume that

y = tan (x)

$y = \tan (x)$ is

f (x) = tan (x)

$f (x) = \tan (x)$ and

y = - tan (\frac{1}{10} x) + 4

$y = - \tan (\frac{1}{10} x) + 4$ is

g (x) = - tan (\frac{x}{10}) + 4

$g (x) = - \tan (\frac{x}{10}) + 4$ .

f (x) = tan (x)

$f (x) = \tan (x)$

g (x) = - tan (\frac{x}{10}) + 4

$g (x) = - \tan (\frac{x}{10}) + 4$

Step 4

Use the form

a tan (b x - c) + d

$a \tan (b x - c) + d$ to find the variables used to find the amplitude, period, phase shift, and vertical shift.

a = - 1

$a = - 1$

b = \frac{1}{10}

$b = \frac{1}{10}$

c = 0

$c = 0$

d = 4

$d = 4$

Step 5

Since the graph of the function

t a n

$t a n$ does not have a maximum or minimum value, there can be no value for the amplitude.

Amplitude: None

Step 6

Find the period using the formula

\frac{π}{| b |}

$\frac{π}{| b |}$ .

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Step 6.1

Find the period of

- tan (\frac{x}{10})

$- \tan (\frac{x}{10})$ .

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Step 6.1.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 6.1.2

Replace

b

$b$ with

\frac{1}{10}

$\frac{1}{10}$ in the formula for period.

\frac{π}{∣ ∣ \frac{1}{10} ∣ ∣}

$\frac{π}{| \frac{1}{10} |}$

Step 6.1.3

\frac{1}{10}

$\frac{1}{10}$ is approximately

0.1

$0.1$ which is positive so remove the absolute value

\frac{π}{\frac{1}{10}}

$\frac{π}{\frac{1}{10}}$

Step 6.1.4

Multiply the numerator by the reciprocal of the denominator.

π \cdot 10

$π \cdot 10$

Step 6.1.5

Move

10

$10$ to the left of

π

$π$ .

10 π

$10 π$

10 π

$10 π$

Step 6.2

Find the period of

4

$4$ .

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Step 6.2.1

The period of the function can be calculated using

\frac{π}{| b |}

$\frac{π}{| b |}$ .

\frac{π}{| b |}

$\frac{π}{| b |}$

Step 6.2.2

Replace

b

$b$ with

\frac{1}{10}

$\frac{1}{10}$ in the formula for period.

\frac{π}{∣ ∣ \frac{1}{10} ∣ ∣}

$\frac{π}{| \frac{1}{10} |}$

Step 6.2.3

\frac{1}{10}

$\frac{1}{10}$ is approximately

0.1

$0.1$ which is positive so remove the absolute value

\frac{π}{\frac{1}{10}}

$\frac{π}{\frac{1}{10}}$

Step 6.2.4

Multiply the numerator by the reciprocal of the denominator.

π \cdot 10

$π \cdot 10$

Step 6.2.5

Move

10

$10$ to the left of

π

$π$ .

10 π

$10 π$

10 π

$10 π$

Step 6.3

The period of addition/subtraction of trig functions is the maximum of the individual periods.

10 π

$10 π$

10 π

$10 π$

Step 7

Find the phase shift using the formula

\frac{c}{b}

$\frac{c}{b}$ .

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Step 7.1

The phase shift of the function can be calculated from

\frac{c}{b}

$\frac{c}{b}$ .

Phase Shift:

\frac{c}{b}

$\frac{c}{b}$

Step 7.2

Replace the values of

c

$c$ and

b

$b$ in the equation for phase shift.

Phase Shift:

\frac{0}{\frac{1}{10}}

$\frac{0}{\frac{1}{10}}$

Step 7.3

Multiply the numerator by the reciprocal of the denominator.

Phase Shift:

0 \cdot 10

$0 \cdot 10$

Step 7.4

Multiply

0

$0$ by

10

$10$ .

Phase Shift:

0

$0$

Phase Shift:

0

$0$

Step 8

List the properties of the trigonometric function.

Amplitude: None

Period:

10 π

$10 π$

Phase Shift: None

Vertical Shift:

4

$4$

Step 9