Trigonometry Examples

y = sin (x) - 2

$y = \sin (x) - 2$

Step 1

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 = - 2

$d = - 2$

Step 2

Find the amplitude

| a |

$| a |$ .

Amplitude:

1

$1$

Step 3

Find the period using the formula

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

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Step 3.1

Find the period of

sin (x)

$\sin (x)$ .

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Step 3.1.1

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 3.1.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 3.1.3

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 3.1.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 3.2

Find the period of

- 2

$- 2$ .

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Step 3.2.1

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 3.2.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 3.2.3

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 3.2.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 3.3

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

2 π

$2 π$

2 π

$2 π$

Step 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{0}{1}

$\frac{0}{1}$

Step 4.3

Divide

0

$0$ by

1

$1$ .

Phase Shift:

0

$0$

Phase Shift:

0

$0$

Step 5

List the properties of the trigonometric function.

Amplitude:

1

$1$

Period:

2 π

$2 π$

Phase Shift: None

Vertical Shift:

- 2

$- 2$

Step 6

Select a few points to graph.

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Step 6.1

Find the point at

x = 0

$x = 0$ .

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Step 6.1.1

Replace the variable

x

$x$ with

0

$0$ in the expression.

f (0) = sin (0) - 2

$f (0) = \sin (0) - 2$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

The exact value of

sin (0)

$\sin (0)$ is

0

$0$ .

f (0) = 0 - 2

$f (0) = 0 - 2$

Step 6.1.2.2

Subtract

2

$2$ from

0

$0$ .

f (0) = - 2

$f (0) = - 2$

Step 6.1.2.3

The final answer is

- 2

$- 2$ .

- 2

$- 2$

- 2

$- 2$

- 2

$- 2$

Step 6.2

Find the point at

x = \frac{π}{2}

$x = \frac{π}{2}$ .

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Step 6.2.1

Replace the variable

x

$x$ with

\frac{π}{2}

$\frac{π}{2}$ in the expression.

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

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

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

The exact value of

sin (\frac{π}{2})

$\sin (\frac{π}{2})$ is

1

$1$ .

f (\frac{π}{2}) = 1 - 2

$f (\frac{π}{2}) = 1 - 2$

Step 6.2.2.2

Subtract

2

$2$ from

1

$1$ .

f (\frac{π}{2}) = - 1

$f (\frac{π}{2}) = - 1$

Step 6.2.2.3

The final answer is

- 1

$- 1$ .

- 1

$- 1$

- 1

$- 1$

- 1

$- 1$

Step 6.3

Find the point at

x = π

$x = π$ .

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Step 6.3.1

Replace the variable

x

$x$ with

π

$π$ in the expression.

f (π) = sin (π) - 2

$f (π) = \sin (π) - 2$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Simplify each term.

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Step 6.3.2.1.1

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

f (π) = sin (0) - 2

$f (π) = \sin (0) - 2$

Step 6.3.2.1.2

The exact value of

sin (0)

$\sin (0)$ is

0

$0$ .

f (π) = 0 - 2

$f (π) = 0 - 2$

f (π) = 0 - 2

$f (π) = 0 - 2$

Step 6.3.2.2

Subtract

2

$2$ from

0

$0$ .

f (π) = - 2

$f (π) = - 2$

Step 6.3.2.3

The final answer is

- 2

$- 2$ .

- 2

$- 2$

- 2

$- 2$

- 2

$- 2$

Step 6.4

Find the point at

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$ .

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Step 6.4.1

Replace the variable

x

$x$ with

\frac{3 π}{2}

$\frac{3 π}{2}$ in the expression.

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

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

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

Simplify each term.

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Step 6.4.2.1.1

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}) = - sin (\frac{π}{2}) - 2

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

Step 6.4.2.1.2

The exact value of

sin (\frac{π}{2})

$\sin (\frac{π}{2})$ is

1

$1$ .

f (\frac{3 π}{2}) = - 1 \cdot 1 - 2

$f (\frac{3 π}{2}) = - 1 \cdot 1 - 2$

Step 6.4.2.1.3

Multiply

- 1

$- 1$ by

1

$1$ .

f (\frac{3 π}{2}) = - 1 - 2

$f (\frac{3 π}{2}) = - 1 - 2$

f (\frac{3 π}{2}) = - 1 - 2

$f (\frac{3 π}{2}) = - 1 - 2$

Step 6.4.2.2

Subtract

2

$2$ from

- 1

$- 1$ .

f (\frac{3 π}{2}) = - 3

$f (\frac{3 π}{2}) = - 3$

Step 6.4.2.3

The final answer is

- 3

$- 3$ .

- 3

$- 3$

- 3

$- 3$

- 3

$- 3$

Step 6.5

Find the point at

x = 2 π

$x = 2 π$ .

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Step 6.5.1

Replace the variable

x

$x$ with

2 π

$2 π$ in the expression.

f (2 π) = sin (2 π) - 2

$f (2 π) = \sin (2 π) - 2$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Simplify each term.

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Step 6.5.2.1.1

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) = sin (0) - 2

$f (2 π) = \sin (0) - 2$

Step 6.5.2.1.2

The exact value of

sin (0)

$\sin (0)$ is

0

$0$ .

f (2 π) = 0 - 2

$f (2 π) = 0 - 2$

f (2 π) = 0 - 2

$f (2 π) = 0 - 2$

Step 6.5.2.2

Subtract

2

$2$ from

0

$0$ .

f (2 π) = - 2

$f (2 π) = - 2$

Step 6.5.2.3

The final answer is

- 2

$- 2$ .

- 2

$- 2$

- 2

$- 2$

- 2

$- 2$

Step 6.6

List the points in a table.

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

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

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

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

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: None

Vertical Shift:

$- 2$

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

Step 8