Trigonometry Examples

y = 4 - sin (x)

$y = 4 - \sin (x)$

Step 1

Rewrite the expression as

- sin (x) + 4

$- \sin (x) + 4$ .

- sin (x) + 4

$- \sin (x) + 4$

Step 2

Use the form

a sin (b x - c) + d

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

a = - 1

$a = - 1$

b = 1

$b = 1$

c = 0

$c = 0$

d = 4

$d = 4$

Step 3

Find the amplitude

| a |

$| a |$ .

Amplitude:

1

$1$

Step 4

Find the period using the formula

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

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Step 4.1

Find the period of

- sin (x)

$- \sin (x)$ .

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Step 4.1.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 4.1.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 4.1.3

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 4.1.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 4.2

Find the period of

4

$4$ .

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Step 4.2.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 4.2.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 4.2.3

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 4.2.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 4.3

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

2 π

$2 π$

2 π

$2 π$

Step 5

Find the phase shift using the formula

\frac{c}{b}

$\frac{c}{b}$ .

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Step 5.1

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 5.2

Replace the values of

c

$c$ and

b

$b$ in the equation for phase shift.

Phase Shift:

\frac{0}{1}

$\frac{0}{1}$

Step 5.3

Divide

0

$0$ by

1

$1$ .

Phase Shift:

0

$0$

Phase Shift:

0

$0$

Step 6

List the properties of the trigonometric function.

Amplitude:

1

$1$

Period:

2 π

$2 π$

Phase Shift: None

Vertical Shift:

4

$4$

Step 7

Select a few points to graph.

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Step 7.1

Find the point at

x = 0

$x = 0$ .

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Step 7.1.1

Replace the variable

x

$x$ with

0

$0$ in the expression.

f (0) = 4 - sin (0)

$f (0) = 4 - \sin (0)$

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

The exact value of

sin (0)

$\sin (0)$ is

0

$0$ .

f (0) = 4 - 0

$f (0) = 4 - 0$

Step 7.1.2.1.2

Multiply

- 1

$- 1$ by

0

$0$ .

f (0) = 4 + 0

$f (0) = 4 + 0$

f (0) = 4 + 0

$f (0) = 4 + 0$

Step 7.1.2.2

Add

4

$4$ and

0

$0$ .

f (0) = 4

$f (0) = 4$

Step 7.1.2.3

The final answer is

4

$4$ .

4

$4$

4

$4$

4

$4$

Step 7.2

Find the point at

x = \frac{π}{2}

$x = \frac{π}{2}$ .

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Step 7.2.1

Replace the variable

x

$x$ with

\frac{π}{2}

$\frac{π}{2}$ in the expression.

f (\frac{π}{2}) = 4 - sin (\frac{π}{2})

$f (\frac{π}{2}) = 4 - \sin (\frac{π}{2})$

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

The exact value of

sin (\frac{π}{2})

$\sin (\frac{π}{2})$ is

1

$1$ .

f (\frac{π}{2}) = 4 - 1 \cdot 1

$f (\frac{π}{2}) = 4 - 1 \cdot 1$

Step 7.2.2.1.2

Multiply

- 1

$- 1$ by

1

$1$ .

f (\frac{π}{2}) = 4 - 1

$f (\frac{π}{2}) = 4 - 1$

f (\frac{π}{2}) = 4 - 1

$f (\frac{π}{2}) = 4 - 1$

Step 7.2.2.2

Subtract

1

$1$ from

4

$4$ .

f (\frac{π}{2}) = 3

$f (\frac{π}{2}) = 3$

Step 7.2.2.3

The final answer is

3

$3$ .

3

$3$

3

$3$

3

$3$

Step 7.3

Find the point at

x = π

$x = π$ .

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Step 7.3.1

Replace the variable

x

$x$ with

π

$π$ in the expression.

f (π) = 4 - sin (π)

$f (π) = 4 - \sin (π)$

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

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

f (π) = 4 - sin (0)

$f (π) = 4 - \sin (0)$

Step 7.3.2.1.2

The exact value of

sin (0)

$\sin (0)$ is

0

$0$ .

f (π) = 4 - 0

$f (π) = 4 - 0$

Step 7.3.2.1.3

Multiply

- 1

$- 1$ by

0

$0$ .

f (π) = 4 + 0

$f (π) = 4 + 0$

f (π) = 4 + 0

$f (π) = 4 + 0$

Step 7.3.2.2

Add

4

$4$ and

0

$0$ .

f (π) = 4

$f (π) = 4$

Step 7.3.2.3

The final answer is

4

$4$ .

4

$4$

4

$4$

4

$4$

Step 7.4

Find the point at

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$ .

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Step 7.4.1

Replace the variable

x

$x$ with

\frac{3 π}{2}

$\frac{3 π}{2}$ in the expression.

f (\frac{3 π}{2}) = 4 - sin (\frac{3 π}{2})

$f (\frac{3 π}{2}) = 4 - \sin (\frac{3 π}{2})$

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

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

f (\frac{3 π}{2}) = 4 + sin (\frac{π}{2})

$f (\frac{3 π}{2}) = 4 + \sin (\frac{π}{2})$

Step 7.4.2.1.2

The exact value of

sin (\frac{π}{2})

$\sin (\frac{π}{2})$ is

1

$1$ .

f (\frac{3 π}{2}) = 4 - (- 1 \cdot 1)

$f (\frac{3 π}{2}) = 4 - (- 1 \cdot 1)$

Step 7.4.2.1.3

Multiply

- (- 1 \cdot 1)

$- (- 1 \cdot 1)$ .

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Step 7.4.2.1.3.1

Multiply

- 1

$- 1$ by

1

$1$ .

f (\frac{3 π}{2}) = 4 + 1

$f (\frac{3 π}{2}) = 4 + 1$

Step 7.4.2.1.3.2

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

f (\frac{3 π}{2}) = 4 + 1

$f (\frac{3 π}{2}) = 4 + 1$

f (\frac{3 π}{2}) = 4 + 1

$f (\frac{3 π}{2}) = 4 + 1$

f (\frac{3 π}{2}) = 4 + 1

$f (\frac{3 π}{2}) = 4 + 1$

Step 7.4.2.2

Add

4

$4$ and

1

$1$ .

f (\frac{3 π}{2}) = 5

$f (\frac{3 π}{2}) = 5$

Step 7.4.2.3

The final answer is

5

$5$ .

5

$5$

5

$5$

5

$5$

Step 7.5

Find the point at

x = 2 π

$x = 2 π$ .

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Step 7.5.1

Replace the variable

x

$x$ with

2 π

$2 π$ in the expression.

f (2 π) = 4 - sin (2 π)

$f (2 π) = 4 - \sin (2 π)$

Step 7.5.2

Simplify the result.

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Step 7.5.2.1

Simplify each term.

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Step 7.5.2.1.1

Subtract full rotations of

2 π

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

0

$0$ and less than

2 π

$2 π$ .

f (2 π) = 4 - sin (0)

$f (2 π) = 4 - \sin (0)$

Step 7.5.2.1.2

The exact value of

sin (0)

$\sin (0)$ is

0

$0$ .

f (2 π) = 4 - 0

$f (2 π) = 4 - 0$

Step 7.5.2.1.3

Multiply

- 1

$- 1$ by

0

$0$ .

f (2 π) = 4 + 0

$f (2 π) = 4 + 0$

f (2 π) = 4 + 0

$f (2 π) = 4 + 0$

Step 7.5.2.2

Add

4

$4$ and

0

$0$ .

f (2 π) = 4

$f (2 π) = 4$

Step 7.5.2.3

The final answer is

4

$4$ .

4

$4$

4

$4$

4

$4$

Step 7.6

List the points in a table.

\begin{matrix} x & f (x) 0 & 4 \frac{π}{2} & 3 π & 4 \frac{3 π}{2} & 5 2 π & 4 \end{matrix}

$\begin{array}{c} x & f (x) \\ 0 & 4 \\ \frac{π}{2} & 3 \\ π & 4 \\ \frac{3 π}{2} & 5 \\ 2 π & 4 \end{array}$

\begin{matrix} x & f (x) 0 & 4 \frac{π}{2} & 3 π & 4 \frac{3 π}{2} & 5 2 π & 4 \end{matrix}

$\begin{array}{c} x & f (x) \\ 0 & 4 \\ \frac{π}{2} & 3 \\ π & 4 \\ \frac{3 π}{2} & 5 \\ 2 π & 4 \end{array}$

Step 8

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

Amplitude:

1

$1$

Period:

2 π

$2 π$

Phase Shift: None

Vertical Shift:

4

$4$

\begin{matrix} x & f (x) 0 & 4 \frac{π}{2} & 3 π & 4 \frac{3 π}{2} & 5 2 π & 4 \end{matrix}

$\begin{array}{c} x & f (x) \\ 0 & 4 \\ \frac{π}{2} & 3 \\ π & 4 \\ \frac{3 π}{2} & 5 \\ 2 π & 4 \end{array}$

Step 9