Trigonometry Examples

y = cos (x + 3)

$y = \cos (x + 3)$

Step 1

Use the form

a cos (b x - c) + d

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

a = 1

$a = 1$

b = 1

$b = 1$

c = - 3

$c = - 3$

d = 0

$d = 0$

Step 2

Find the amplitude

| a |

$| a |$ .

Amplitude:

1

$1$

Step 3

Find the period of

cos (x + 3)

$\cos (x + 3)$ .

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Step 3.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 3.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 3.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 3.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 4

Find the phase shift using the formula

\frac{c}{b}

$\frac{c}{b}$ .

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Step 4.1

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 4.2

Replace the values of

c

$c$ and

b

$b$ in the equation for phase shift.

Phase Shift:

\frac{- 3}{1}

$\frac{- 3}{1}$

Step 4.3

Divide

- 3

$- 3$ by

1

$1$ .

Phase Shift:

- 3

$- 3$

Phase Shift:

- 3

$- 3$

Step 5

List the properties of the trigonometric function.

Amplitude:

1

$1$

Period:

2 π

$2 π$

Phase Shift:

- 3

$- 3$ (

3

$3$ 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 = - 3

$x = - 3$ .

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Step 6.1.1

Replace the variable

x

$x$ with

- 3

$- 3$ in the expression.

f (- 3) = cos ((- 3) + 3)

$f (- 3) = \cos ((- 3) + 3)$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Add

- 3

$- 3$ and

3

$3$ .

f (- 3) = cos (0)

$f (- 3) = \cos (0)$

Step 6.1.2.2

The exact value of

cos (0)

$\cos (0)$ is

1

$1$ .

f (- 3) = 1

$f (- 3) = 1$

Step 6.1.2.3

The final answer is

1

$1$ .

1

$1$

1

$1$

1

$1$

Step 6.2

Find the point at

x = \frac{π}{2} - 3

$x = \frac{π}{2} - 3$ .

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Step 6.2.1

Replace the variable

x

$x$ with

\frac{π}{2} - 3

$\frac{π}{2} - 3$ in the expression.

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

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

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

Add

- 3

$- 3$ and

3

$3$ .

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

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

Step 6.2.2.2

Add

\frac{π}{2}

$\frac{π}{2}$ and

0

$0$ .

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

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

Step 6.2.2.3

The exact value of

cos (\frac{π}{2})

$\cos (\frac{π}{2})$ is

0

$0$ .

f (\frac{π}{2} - 3) = 0

$f (\frac{π}{2} - 3) = 0$

Step 6.2.2.4

The final answer is

0

$0$ .

0

$0$

0

$0$

0

$0$

Step 6.3

Find the point at

x = π - 3

$x = π - 3$ .

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Step 6.3.1

Replace the variable

x

$x$ with

π - 3

$π - 3$ in the expression.

f (π - 3) = cos ((π - 3) + 3)

$f (π - 3) = \cos ((π - 3) + 3)$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Add

- 3

$- 3$ and

3

$3$ .

f (π - 3) = cos (π + 0)

$f (π - 3) = \cos (π + 0)$

Step 6.3.2.2

Add

π

$π$ and

0

$0$ .

f (π - 3) = cos (π)

$f (π - 3) = \cos (π)$

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 - 3) = - cos (0)

$f (π - 3) = - \cos (0)$

Step 6.3.2.4

The exact value of

cos (0)

$\cos (0)$ is

1

$1$ .

f (π - 3) = - 1 \cdot 1

$f (π - 3) = - 1 \cdot 1$

Step 6.3.2.5

Multiply

- 1

$- 1$ by

1

$1$ .

f (π - 3) = - 1

$f (π - 3) = - 1$

Step 6.3.2.6

The final answer is

- 1

$- 1$ .

- 1

$- 1$

- 1

$- 1$

- 1

$- 1$

Step 6.4

Find the point at

x = \frac{3 π}{2} - 3

$x = \frac{3 π}{2} - 3$ .

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Step 6.4.1

Replace the variable

x

$x$ with

\frac{3 π}{2} - 3

$\frac{3 π}{2} - 3$ in the expression.

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

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

Add

$- 3$ and

$3$ .

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

Step 6.4.2.2

Add

$\frac{3 π}{2}$ and

$0$ .

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

Step 6.4.2.3

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

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

Step 6.4.2.4

The exact value of

$\cos (\frac{π}{2})$ is

$0$ .

$f (\frac{3 π}{2} - 3) = 0$

Step 6.4.2.5

The final answer is

$0$ .

$0$

Step 6.5

Find the point at

$x = 2 π - 3$ .

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Step 6.5.1

Replace the variable

$x$ with

$2 π - 3$ in the expression.

$f (2 π - 3) = \cos ((2 π - 3) + 3)$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Add

$- 3$ and

$3$ .

$f (2 π - 3) = \cos (2 π + 0)$

Step 6.5.2.2

Add

$2 π$ and

$0$ .

$f (2 π - 3) = \cos (2 π)$

Step 6.5.2.3

Subtract full rotations of

$2 π$ until the angle is greater than or equal to

$0$ and less than

$2 π$ .

$f (2 π - 3) = \cos (0)$

Step 6.5.2.4

The exact value of

$\cos (0)$ is

$1$ .

$f (2 π - 3) = 1$

Step 6.5.2.5

The final answer is

$1$ .

$1$

Step 6.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ - 3 & 1 \\ \frac{π}{2} - 3 & 0 \\ π - 3 & - 1 \\ \frac{3 π}{2} - 3 & 0 \\ 2 π - 3 & 1 \end{array}$

Step 7

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

Amplitude:

$1$

Period:

$2 π$

Phase Shift:

$- 3$ (

$3$ to the left)

Vertical Shift: None

$\begin{array}{c} x & f (x) \\ - 3 & 1 \\ \frac{π}{2} - 3 & 0 \\ π - 3 & - 1 \\ \frac{3 π}{2} - 3 & 0 \\ 2 π - 3 & 1 \end{array}$

Step 8