Trigonometry Examples

$y = \sin (x - 2) - 1$

Step 1

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

$a = 1$

$b = 1$

$c = 2$

$d = - 1$

Step 2

Find the amplitude $| a |$ .

Amplitude: $1$

Step 3

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

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Step 3.1

Find the period of $\sin (x - 2)$ .

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Step 3.1.1

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

$\frac{2 π}{| b |}$

Step 3.1.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.1.3

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

$\frac{2 π}{1}$

Step 3.1.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2

Find the period of $- 1$ .

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Step 3.2.1

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

$\frac{2 π}{| b |}$

Step 3.2.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.2.3

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

$\frac{2 π}{1}$

Step 3.2.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.3

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

$2 π$

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: $1$

Period: $2 π$

Phase Shift: $2$ ( $2$ to the right)

Vertical Shift: $- 1$

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.

$f (2) = \sin ((2) - 2) - 1$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Simplify each term.

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Step 6.1.2.1.1

Subtract $2$ from $2$ .

$f (2) = \sin (0) - 1$

Step 6.1.2.1.2

The exact value of $\sin (0)$ is $0$ .

$f (2) = 0 - 1$

Step 6.1.2.2

Subtract $1$ from $0$ .

$f (2) = - 1$

Step 6.1.2.3

The final answer is $- 1$ .

$- 1$

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.

$f (\frac{π}{2} + 2) = \sin ((\frac{π}{2} + 2) - 2) - 1$

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

Simplify each term.

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Step 6.2.2.1.1

Subtract $2$ from $2$ .

$f (\frac{π}{2} + 2) = \sin (\frac{π}{2} + 0) - 1$

Step 6.2.2.1.2

Add $\frac{π}{2}$ and $0$ .

$f (\frac{π}{2} + 2) = \sin (\frac{π}{2}) - 1$

Step 6.2.2.1.3

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{π}{2} + 2) = 1 - 1$

Step 6.2.2.2

Subtract $1$ from $1$ .

$f (\frac{π}{2} + 2) = 0$

Step 6.2.2.3

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.

$f (π + 2) = \sin ((π + 2) - 2) - 1$

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

Subtract $2$ from $2$ .

$f (π + 2) = \sin (π + 0) - 1$

Step 6.3.2.1.2

Add $π$ and $0$ .

$f (π + 2) = \sin (π) - 1$

Step 6.3.2.1.3

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

$f (π + 2) = \sin (0) - 1$

Step 6.3.2.1.4

The exact value of $\sin (0)$ is $0$ .

$f (π + 2) = 0 - 1$

Step 6.3.2.2

Subtract $1$ from $0$ .

$f (π + 2) = - 1$

Step 6.3.2.3

The final answer is $- 1$ .

$- 1$

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.

$f (\frac{3 π}{2} + 2) = \sin ((\frac{3 π}{2} + 2) - 2) - 1$

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

Subtract $2$ from $2$ .

$f (\frac{3 π}{2} + 2) = \sin (\frac{3 π}{2} + 0) - 1$

Step 6.4.2.1.2

Add $\frac{3 π}{2}$ and $0$ .

$f (\frac{3 π}{2} + 2) = \sin (\frac{3 π}{2}) - 1$

Step 6.4.2.1.3

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.

$f (\frac{3 π}{2} + 2) = - \sin (\frac{π}{2}) - 1$

Step 6.4.2.1.4

The exact value of $\sin (\frac{π}{2})$ is $1$ .

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

Step 6.4.2.1.5

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{2} + 2) = - 1 - 1$

Step 6.4.2.2

Subtract $1$ from $- 1$ .

$f (\frac{3 π}{2} + 2) = - 2$

Step 6.4.2.3

The final answer is $- 2$ .

$- 2$

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.

$f (2 π + 2) = \sin ((2 π + 2) - 2) - 1$

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 $2$ from $2$ .

$f (2 π + 2) = \sin (2 π + 0) - 1$

Step 6.5.2.1.2

Add $2 π$ and $0$ .

$f (2 π + 2) = \sin (2 π) - 1$

Step 6.5.2.1.3

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

$f (2 π + 2) = \sin (0) - 1$

Step 6.5.2.1.4

The exact value of $\sin (0)$ is $0$ .

$f (2 π + 2) = 0 - 1$

Step 6.5.2.2

Subtract $1$ from $0$ .

$f (2 π + 2) = - 1$

Step 6.5.2.3

The final answer is $- 1$ .

$- 1$

Step 6.6

List the points in a table.

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

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: $2$ ( $2$ to the right)

Vertical Shift: $- 1$

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

Step 8