Trigonometry Examples

$h = 40 - 30 \cos (18 t)$

Step 1

Rewrite the expression as $- 30 \cos (18 x) + 40$ .

$- 30 \cos (18 x) + 40$

Step 2

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

$a = - 30$

$b = 18$

$c = 0$

$d = 40$

Step 3

Find the amplitude $| a |$ .

Amplitude: $30$

Step 4

Find the period using the formula $\frac{2 π}{| b |}$ .

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Step 4.1

Find the period of $- 30 \cos (18 x)$ .

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Step 4.1.1

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

$\frac{2 π}{| b |}$

Step 4.1.2

Replace $b$ with $18$ in the formula for period.

$\frac{2 π}{| 18 |}$

Step 4.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $18$ is $18$ .

$\frac{2 π}{18}$

Step 4.1.4

Cancel the common factor of $2$ and $18$ .

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Step 4.1.4.1

Factor $2$ out of $2 π$ .

$\frac{2 (π)}{18}$

Step 4.1.4.2

Cancel the common factors.

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Step 4.1.4.2.1

Factor $2$ out of $18$ .

$\frac{2 π}{2 \cdot 9}$

Step 4.1.4.2.2

Cancel the common factor.

$\frac{2 π}{2 \cdot 9}$

Step 4.1.4.2.3

Rewrite the expression.

$\frac{π}{9}$

Step 4.2

Find the period of $40$ .

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Step 4.2.1

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

$\frac{2 π}{| b |}$

Step 4.2.2

Replace $b$ with $18$ in the formula for period.

$\frac{2 π}{| 18 |}$

Step 4.2.3

The absolute value is the distance between a number and zero. The distance between $0$ and $18$ is $18$ .

$\frac{2 π}{18}$

Step 4.2.4

Cancel the common factor of $2$ and $18$ .

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Step 4.2.4.1

Factor $2$ out of $2 π$ .

$\frac{2 (π)}{18}$

Step 4.2.4.2

Cancel the common factors.

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Step 4.2.4.2.1

Factor $2$ out of $18$ .

$\frac{2 π}{2 \cdot 9}$

Step 4.2.4.2.2

Cancel the common factor.

$\frac{2 π}{2 \cdot 9}$

Step 4.2.4.2.3

Rewrite the expression.

$\frac{π}{9}$

Step 4.3

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

$\frac{π}{9}$

Step 5

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

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Step 5.1

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

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

Step 5.2

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

Phase Shift: $\frac{0}{18}$

Step 5.3

Divide $0$ by $18$ .

Phase Shift: $0$

Step 6

List the properties of the trigonometric function.

Amplitude: $30$

Period: $\frac{π}{9}$

Phase Shift: None

Vertical Shift: $40$

Step 7

Select a few points to graph.

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Step 7.1

Find the point at $x = 0$ .

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Step 7.1.1

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

$f (0) = 40 - 30 \cos (18 (0))$

Step 7.1.2

Simplify the result.

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Step 7.1.2.1

Simplify each term.

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Step 7.1.2.1.1

Multiply $18$ by $0$ .

$f (0) = 40 - 30 \cos (0)$

Step 7.1.2.1.2

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

$f (0) = 40 - 30 \cdot 1$

Step 7.1.2.1.3

Multiply $- 30$ by $1$ .

$f (0) = 40 - 30$

Step 7.1.2.2

Subtract $30$ from $40$ .

$f (0) = 10$

Step 7.1.2.3

The final answer is $10$ .

$10$

Step 7.2

Find the point at $x = \frac{π}{36}$ .

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Step 7.2.1

Replace the variable $x$ with $\frac{π}{36}$ in the expression.

$f (\frac{π}{36}) = 40 - 30 \cos (18 (\frac{π}{36}))$

Step 7.2.2

Simplify the result.

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Step 7.2.2.1

Simplify each term.

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Step 7.2.2.1.1

Cancel the common factor of $18$ .

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Step 7.2.2.1.1.1

Factor $18$ out of $36$ .

$f (\frac{π}{36}) = 40 - 30 \cos (18 (\frac{π}{18 (2)}))$

Step 7.2.2.1.1.2

Cancel the common factor.

$f (\frac{π}{36}) = 40 - 30 \cos (18 (\frac{π}{18 \cdot 2}))$

Step 7.2.2.1.1.3

Rewrite the expression.

$f (\frac{π}{36}) = 40 - 30 \cos (\frac{π}{2})$

Step 7.2.2.1.2

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

$f (\frac{π}{36}) = 40 - 30 \cdot 0$

Step 7.2.2.1.3

Multiply $- 30$ by $0$ .

$f (\frac{π}{36}) = 40 + 0$

Step 7.2.2.2

Add $40$ and $0$ .

$f (\frac{π}{36}) = 40$

Step 7.2.2.3

The final answer is $40$ .

$40$

Step 7.3

Find the point at $x = \frac{π}{9}$ .

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Step 7.3.1

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

$f (\frac{π}{9}) = 40 - 30 \cos (18 (\frac{π}{9}))$

Step 7.3.2

Simplify the result.

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Step 7.3.2.1

Simplify each term.

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Step 7.3.2.1.1

Cancel the common factor of $9$ .

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Step 7.3.2.1.1.1

Factor $9$ out of $18$ .

$f (\frac{π}{9}) = 40 - 30 \cos (9 (2) (\frac{π}{9}))$

Step 7.3.2.1.1.2

Cancel the common factor.

$f (\frac{π}{9}) = 40 - 30 \cos (9 \cdot (2 (\frac{π}{9})))$

Step 7.3.2.1.1.3

Rewrite the expression.

$f (\frac{π}{9}) = 40 - 30 \cos (2 π)$

Step 7.3.2.1.2

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

$f (\frac{π}{9}) = 40 - 30 \cos (0)$

Step 7.3.2.1.3

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

$f (\frac{π}{9}) = 40 - 30 \cdot 1$

Step 7.3.2.1.4

Multiply $- 30$ by $1$ .

$f (\frac{π}{9}) = 40 - 30$

Step 7.3.2.2

Subtract $30$ from $40$ .

$f (\frac{π}{9}) = 10$

Step 7.3.2.3

The final answer is $10$ .

$10$

Step 7.4

Find the point at $x = \frac{5 π}{36}$ .

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Step 7.4.1

Replace the variable $x$ with $\frac{5 π}{36}$ in the expression.

$f (\frac{5 π}{36}) = 40 - 30 \cos (18 (\frac{5 π}{36}))$

Step 7.4.2

Simplify the result.

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Step 7.4.2.1

Simplify each term.

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Step 7.4.2.1.1

Cancel the common factor of $18$ .

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Step 7.4.2.1.1.1

Factor $18$ out of $36$ .

$f (\frac{5 π}{36}) = 40 - 30 \cos (18 (\frac{5 π}{18 (2)}))$

Step 7.4.2.1.1.2

Cancel the common factor.

$f (\frac{5 π}{36}) = 40 - 30 \cos (18 (\frac{5 π}{18 \cdot 2}))$

Step 7.4.2.1.1.3

Rewrite the expression.

$f (\frac{5 π}{36}) = 40 - 30 \cos (\frac{5 π}{2})$

Step 7.4.2.1.2

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

$f (\frac{5 π}{36}) = 40 - 30 \cos (\frac{π}{2})$

Step 7.4.2.1.3

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

$f (\frac{5 π}{36}) = 40 - 30 \cdot 0$

Step 7.4.2.1.4

Multiply $- 30$ by $0$ .

$f (\frac{5 π}{36}) = 40 + 0$

Step 7.4.2.2

Add $40$ and $0$ .

$f (\frac{5 π}{36}) = 40$

Step 7.4.2.3

The final answer is $40$ .

$40$

Step 7.5

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 10 \\ \frac{π}{36} & 40 \\ \frac{π}{12} & 40 \\ \frac{π}{9} & 10 \\ \frac{5 π}{36} & 40 \end{array}$

Step 8

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

Amplitude: $30$

Period: $\frac{π}{9}$

Phase Shift: None

Vertical Shift: $40$

$\begin{array}{c} x & f (x) \\ 0 & 10 \\ \frac{π}{36} & 40 \\ \frac{π}{12} & 40 \\ \frac{π}{9} & 10 \\ \frac{5 π}{36} & 40 \end{array}$

Step 9