Trigonometry Examples

$f (x) = 9 \cos (9 x)$

Step 1

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

$a = 9$

$b = 9$

$c = 0$

$d = 0$

Step 2

Find the amplitude $| a |$ .

Amplitude: $9$

Step 3

Find the period of $9 \cos (9 x)$ .

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Step 3.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.2

Replace $b$ with $9$ in the formula for period.

$\frac{2 π}{| 9 |}$

Step 3.3

The absolute value is the distance between a number and zero. The distance between $0$ and $9$ is $9$ .

$\frac{2 π}{9}$

Step 4

Find the phase shift using the formula $\frac{c}{b}$ .

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Step 4.1

The phase shift of the function can be calculated from $\frac{c}{b}$ .

Phase Shift: $\frac{c}{b}$

Step 4.2

Replace the values of $c$ and $b$ in the equation for phase shift.

Phase Shift: $\frac{0}{9}$

Step 4.3

Divide $0$ by $9$ .

Phase Shift: $0$

Step 5

List the properties of the trigonometric function.

Amplitude: $9$

Period: $\frac{2 π}{9}$

Phase Shift: None

Vertical Shift: None

Step 6

Select a few points to graph.

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Step 6.1

Find the point at $x = 0$ .

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Step 6.1.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = 9 \cos (9 (0))$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Multiply $9$ by $0$ .

$f (0) = 9 \cos (0)$

Step 6.1.2.2

The exact value of $\cos (0)$ is $1$ .

$f (0) = 9 \cdot 1$

Step 6.1.2.3

Multiply $9$ by $1$ .

$f (0) = 9$

Step 6.1.2.4

The final answer is $9$ .

$9$

Step 6.2

Find the point at $x = \frac{π}{18}$ .

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Step 6.2.1

Replace the variable $x$ with $\frac{π}{18}$ in the expression.

$f (\frac{π}{18}) = 9 \cos (9 (\frac{π}{18}))$

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

Cancel the common factor of $9$ .

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Step 6.2.2.1.1

Factor $9$ out of $18$ .

$f (\frac{π}{18}) = 9 \cos (9 (\frac{π}{9 (2)}))$

Step 6.2.2.1.2

Cancel the common factor.

$f (\frac{π}{18}) = 9 \cos (9 (\frac{π}{9 \cdot 2}))$

Step 6.2.2.1.3

Rewrite the expression.

$f (\frac{π}{18}) = 9 \cos (\frac{π}{2})$

Step 6.2.2.2

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$f (\frac{π}{18}) = 9 \cdot 0$

Step 6.2.2.3

Multiply $9$ by $0$ .

$f (\frac{π}{18}) = 0$

Step 6.2.2.4

The final answer is $0$ .

$0$

Step 6.3

Find the point at $x = \frac{π}{9}$ .

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Step 6.3.1

Replace the variable $x$ with $\frac{π}{9}$ in the expression.

$f (\frac{π}{9}) = 9 \cos (9 (\frac{π}{9}))$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Cancel the common factor of $9$ .

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Step 6.3.2.1.1

Cancel the common factor.

$f (\frac{π}{9}) = 9 \cos (9 (\frac{π}{9}))$

Step 6.3.2.1.2

Rewrite the expression.

$f (\frac{π}{9}) = 9 \cos (π)$

Step 6.3.2.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$f (\frac{π}{9}) = 9 (- \cos (0))$

Step 6.3.2.3

The exact value of $\cos (0)$ is $1$ .

$f (\frac{π}{9}) = 9 (- 1 \cdot 1)$

Step 6.3.2.4

Multiply $9 (- 1 \cdot 1)$ .

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Step 6.3.2.4.1

Multiply $- 1$ by $1$ .

$f (\frac{π}{9}) = 9 \cdot - 1$

Step 6.3.2.4.2

Multiply $9$ by $- 1$ .

$f (\frac{π}{9}) = - 9$

Step 6.3.2.5

The final answer is $- 9$ .

$- 9$

Step 6.4

Find the point at $x = \frac{π}{6}$ .

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Step 6.4.1

Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = 9 \cos (9 (\frac{π}{6}))$

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

Cancel the common factor of $3$ .

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Step 6.4.2.1.1

Factor $3$ out of $9$ .

$f (\frac{π}{6}) = 9 \cos (3 (3) (\frac{π}{6}))$

Step 6.4.2.1.2

Factor $3$ out of $6$ .

$f (\frac{π}{6}) = 9 \cos (3 \cdot (3 (\frac{π}{3 \cdot 2})))$

Step 6.4.2.1.3

Cancel the common factor.

$f (\frac{π}{6}) = 9 \cos (3 \cdot (3 (\frac{π}{3 \cdot 2})))$

Step 6.4.2.1.4

Rewrite the expression.

$f (\frac{π}{6}) = 9 \cos (3 (\frac{π}{2}))$

Step 6.4.2.2

Combine $3$ and $\frac{π}{2}$ .

$f (\frac{π}{6}) = 9 \cos (\frac{3 π}{2})$

Step 6.4.2.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$f (\frac{π}{6}) = 9 \cos (\frac{π}{2})$

Step 6.4.2.4

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$f (\frac{π}{6}) = 9 \cdot 0$

Step 6.4.2.5

Multiply $9$ by $0$ .

$f (\frac{π}{6}) = 0$

Step 6.4.2.6

The final answer is $0$ .

$0$

Step 6.5

Find the point at $x = \frac{2 π}{9}$ .

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Step 6.5.1

Replace the variable $x$ with $\frac{2 π}{9}$ in the expression.

$f (\frac{2 π}{9}) = 9 \cos (9 (\frac{2 π}{9}))$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Cancel the common factor of $9$ .

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Step 6.5.2.1.1

Cancel the common factor.

$f (\frac{2 π}{9}) = 9 \cos (9 (\frac{2 π}{9}))$

Step 6.5.2.1.2

Rewrite the expression.

$f (\frac{2 π}{9}) = 9 \cos (2 π)$

Step 6.5.2.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$f (\frac{2 π}{9}) = 9 \cos (0)$

Step 6.5.2.3

The exact value of $\cos (0)$ is $1$ .

$f (\frac{2 π}{9}) = 9 \cdot 1$

Step 6.5.2.4

Multiply $9$ by $1$ .

$f (\frac{2 π}{9}) = 9$

Step 6.5.2.5

The final answer is $9$ .

$9$

Step 6.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 9 \\ \frac{π}{18} & 0 \\ \frac{π}{9} & - 9 \\ \frac{π}{6} & 0 \\ \frac{2 π}{9} & 9 \end{array}$

Step 7

The trig function can be graphed using the amplitude, period, phase shift, vertical shift, and the points.

Amplitude: $9$

Period: $\frac{2 π}{9}$

Phase Shift: None

Vertical Shift: None

$\begin{array}{c} x & f (x) \\ 0 & 9 \\ \frac{π}{18} & 0 \\ \frac{π}{9} & - 9 \\ \frac{π}{6} & 0 \\ \frac{2 π}{9} & 9 \end{array}$

Step 8