Trigonometry Examples

$y = 3 \sin (x - π)$

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.

$a = 3$

$b = 1$

$c = π$

$d = 0$

Step 2

Find the amplitude $| a |$ .

Amplitude: $3$

Step 3

Find the period of $3 \sin (x - π)$ .

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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: $\frac{π}{1}$

Step 4.3

Divide $π$ by $1$ .

Phase Shift: $π$

Step 5

List the properties of the trigonometric function.

Amplitude: $3$

Period: $2 π$

Phase Shift: $π$ ( $π$ to the right)

Vertical Shift: None

Step 6

Select a few points to graph.

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Step 6.1

Find the point at $x = π$ .

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Step 6.1.1

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

$f (π) = 3 \sin ((π) - π)$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Subtract $π$ from $π$ .

$f (π) = 3 \sin (0)$

Step 6.1.2.2

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

$f (π) = 3 \cdot 0$

Step 6.1.2.3

Multiply $3$ by $0$ .

$f (π) = 0$

Step 6.1.2.4

The final answer is $0$ .

$0$

Step 6.2

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

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Step 6.2.1

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

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

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

To write $- π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 6.2.2.2

Combine fractions.

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Step 6.2.2.2.1

Combine $- π$ and $\frac{2}{2}$ .

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

Step 6.2.2.2.2

Combine the numerators over the common denominator.

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

Step 6.2.2.3

Simplify the numerator.

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Step 6.2.2.3.1

Multiply $2$ by $- 1$ .

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

Step 6.2.2.3.2

Subtract $2 π$ from $3 π$ .

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

Step 6.2.2.4

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

$f (\frac{3 π}{2}) = 3 \cdot 1$

Step 6.2.2.5

Multiply $3$ by $1$ .

$f (\frac{3 π}{2}) = 3$

Step 6.2.2.6

The final answer is $3$ .

$3$

Step 6.3

Find the point at $x = 2 π$ .

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Step 6.3.1

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

$f (2 π) = 3 \sin ((2 π) - π)$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Subtract $π$ from $2 π$ .

$f (2 π) = 3 \sin (π)$

Step 6.3.2.2

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

$f (2 π) = 3 \sin (0)$

Step 6.3.2.3

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

$f (2 π) = 3 \cdot 0$

Step 6.3.2.4

Multiply $3$ by $0$ .

$f (2 π) = 0$

Step 6.3.2.5

The final answer is $0$ .

$0$

Step 6.4

Find the point at $x = \frac{5 π}{2}$ .

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Step 6.4.1

Replace the variable $x$ with $\frac{5 π}{2}$ in the expression.

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

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

To write $- π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 6.4.2.2

Combine fractions.

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Step 6.4.2.2.1

Combine $- π$ and $\frac{2}{2}$ .

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

Step 6.4.2.2.2

Combine the numerators over the common denominator.

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

Step 6.4.2.3

Simplify the numerator.

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Step 6.4.2.3.1

Multiply $2$ by $- 1$ .

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

Step 6.4.2.3.2

Subtract $2 π$ from $5 π$ .

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

Step 6.4.2.4

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.

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

Step 6.4.2.5

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

$f (\frac{5 π}{2}) = 3 (- 1 \cdot 1)$

Step 6.4.2.6

Multiply $3 (- 1 \cdot 1)$ .

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Step 6.4.2.6.1

Multiply $- 1$ by $1$ .

$f (\frac{5 π}{2}) = 3 \cdot - 1$

Step 6.4.2.6.2

Multiply $3$ by $- 1$ .

$f (\frac{5 π}{2}) = - 3$

Step 6.4.2.7

The final answer is $- 3$ .

$- 3$

Step 6.5

Find the point at $x = 3 π$ .

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Step 6.5.1

Replace the variable $x$ with $3 π$ in the expression.

$f (3 π) = 3 \sin ((3 π) - π)$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Subtract $π$ from $3 π$ .

$f (3 π) = 3 \sin (2 π)$

Step 6.5.2.2

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

$f (3 π) = 3 \sin (0)$

Step 6.5.2.3

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

$f (3 π) = 3 \cdot 0$

Step 6.5.2.4

Multiply $3$ by $0$ .

$f (3 π) = 0$

Step 6.5.2.5

The final answer is $0$ .

$0$

Step 6.6

List the points in a table.

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

Step 7

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

Amplitude: $3$

Period: $2 π$

Phase Shift: $π$ ( $π$ to the right)

Vertical Shift: None

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

Step 8