Trigonometry Examples

$y = 5 \cos (x + 2)$

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.

$a = 5$

$b = 1$

$c = - 2$

$d = 0$

Step 2

Find the amplitude $| a |$ .

Amplitude: $5$

Step 3

Find the period of $5 \cos (x + 2)$ .

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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 $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.3

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

$\frac{2 π}{1}$

Step 3.4

Divide $2 π$ by $1$ .

$2 π$

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: $\frac{- 2}{1}$

Step 4.3

Divide $- 2$ by $1$ .

Phase Shift: $- 2$

Step 5

List the properties of the trigonometric function.

Amplitude: $5$

Period: $2 π$

Phase Shift: $- 2$ ( $2$ to the left)

Vertical Shift: None

Step 6

Select a few points to graph.

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Step 6.1

Find the point at $x = - 2$ .

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Step 6.1.1

Replace the variable $x$ with $- 2$ in the expression.

$f (- 2) = 5 \cos ((- 2) + 2)$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Add $- 2$ and $2$ .

$f (- 2) = 5 \cos (0)$

Step 6.1.2.2

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

$f (- 2) = 5 \cdot 1$

Step 6.1.2.3

Multiply $5$ by $1$ .

$f (- 2) = 5$

Step 6.1.2.4

The final answer is $5$ .

$5$

Step 6.2

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

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Step 6.2.1

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

$f (\frac{π}{2} - 2) = 5 \cos ((\frac{π}{2} - 2) + 2)$

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

Add $- 2$ and $2$ .

$f (\frac{π}{2} - 2) = 5 \cos (\frac{π}{2} + 0)$

Step 6.2.2.2

Add $\frac{π}{2}$ and $0$ .

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

Step 6.2.2.3

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

$f (\frac{π}{2} - 2) = 5 \cdot 0$

Step 6.2.2.4

Multiply $5$ by $0$ .

$f (\frac{π}{2} - 2) = 0$

Step 6.2.2.5

The final answer is $0$ .

$0$

Step 6.3

Find the point at $x = π - 2$ .

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Step 6.3.1

Replace the variable $x$ with $π - 2$ in the expression.

$f (π - 2) = 5 \cos ((π - 2) + 2)$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Add $- 2$ and $2$ .

$f (π - 2) = 5 \cos (π + 0)$

Step 6.3.2.2

Add $π$ and $0$ .

$f (π - 2) = 5 \cos (π)$

Step 6.3.2.3

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.

$f (π - 2) = 5 (- \cos (0))$

Step 6.3.2.4

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

$f (π - 2) = 5 (- 1 \cdot 1)$

Step 6.3.2.5

Multiply $5 (- 1 \cdot 1)$ .

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Step 6.3.2.5.1

Multiply $- 1$ by $1$ .

$f (π - 2) = 5 \cdot - 1$

Step 6.3.2.5.2

Multiply $5$ by $- 1$ .

$f (π - 2) = - 5$

Step 6.3.2.6

The final answer is $- 5$ .

$- 5$

Step 6.4

Find the point at $x = \frac{3 π}{2} - 2$ .

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Step 6.4.1

Replace the variable $x$ with $\frac{3 π}{2} - 2$ in the expression.

$f (\frac{3 π}{2} - 2) = 5 \cos ((\frac{3 π}{2} - 2) + 2)$

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

Add $- 2$ and $2$ .

$f (\frac{3 π}{2} - 2) = 5 \cos (\frac{3 π}{2} + 0)$

Step 6.4.2.2

Add $\frac{3 π}{2}$ and $0$ .

$f (\frac{3 π}{2} - 2) = 5 \cos (\frac{3 π}{2})$

Step 6.4.2.3

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

$f (\frac{3 π}{2} - 2) = 5 \cos (\frac{π}{2})$

Step 6.4.2.4

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

$f (\frac{3 π}{2} - 2) = 5 \cdot 0$

Step 6.4.2.5

Multiply $5$ by $0$ .

$f (\frac{3 π}{2} - 2) = 0$

Step 6.4.2.6

The final answer is $0$ .

$0$

Step 6.5

Find the point at $x = 2 π - 2$ .

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Step 6.5.1

Replace the variable $x$ with $2 π - 2$ in the expression.

$f (2 π - 2) = 5 \cos ((2 π - 2) + 2)$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Add $- 2$ and $2$ .

$f (2 π - 2) = 5 \cos (2 π + 0)$

Step 6.5.2.2

Add $2 π$ and $0$ .

$f (2 π - 2) = 5 \cos (2 π)$

Step 6.5.2.3

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

$f (2 π - 2) = 5 \cos (0)$

Step 6.5.2.4

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

$f (2 π - 2) = 5 \cdot 1$

Step 6.5.2.5

Multiply $5$ by $1$ .

$f (2 π - 2) = 5$

Step 6.5.2.6

The final answer is $5$ .

$5$

Step 6.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ - 2 & 5 \\ \frac{π}{2} - 2 & 0 \\ π - 2 & - 5 \\ \frac{3 π}{2} - 2 & 0 \\ 2 π - 2 & 5 \end{array}$

Step 7

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

Amplitude: $5$

Period: $2 π$

Phase Shift: $- 2$ ( $2$ to the left)

Vertical Shift: None

$\begin{array}{c} x & f (x) \\ - 2 & 5 \\ \frac{π}{2} - 2 & 0 \\ π - 2 & - 5 \\ \frac{3 π}{2} - 2 & 0 \\ 2 π - 2 & 5 \end{array}$

Step 8