Trigonometry Examples

$5 \sin (\frac{π x}{2})$

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.

$a = 5$

$b = \frac{π}{2}$

$c = 0$

$d = 0$

Step 2

Find the amplitude $| a |$ .

Amplitude: $5$

Step 3

Find the period of $5 \sin (\frac{π x}{2})$ .

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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 $\frac{π}{2}$ in the formula for period.

$\frac{2 π}{| \frac{π}{2} |}$

Step 3.3

$\frac{π}{2}$ is approximately $1.57079632$ which is positive so remove the absolute value

$\frac{2 π}{\frac{π}{2}}$

Step 3.4

Multiply the numerator by the reciprocal of the denominator.

$2 π \frac{2}{π}$

Step 3.5

Cancel the common factor of $π$ .

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Step 3.5.1

Factor $π$ out of $2 π$ .

$π \cdot 2 \frac{2}{π}$

Step 3.5.2

Cancel the common factor.

$π \cdot 2 \frac{2}{π}$

Step 3.5.3

Rewrite the expression.

$2 \cdot 2$

Step 3.6

Multiply $2$ by $2$ .

$4$

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: $\frac{0}{\frac{π}{2}}$

Step 4.3

Multiply the numerator by the reciprocal of the denominator.

Phase Shift: $0 (\frac{2}{π})$

Step 4.4

Multiply $0$ by $\frac{2}{π}$ .

Phase Shift: $0$

Step 5

List the properties of the trigonometric function.

Amplitude: $5$

Period: $4$

Phase Shift: None

Vertical Shift: None

Step 6

Select a few points to graph.

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Step 6.1

Find the point at $x = 0$ .

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Step 6.1.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = 5 \sin (\frac{π (0)}{2})$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Cancel the common factor of $0$ and $2$ .

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Step 6.1.2.1.1

Factor $2$ out of $π (0)$ .

$f (0) = 5 \sin (\frac{2 (π \cdot (0))}{2})$

Step 6.1.2.1.2

Cancel the common factors.

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Step 6.1.2.1.2.1

Factor $2$ out of $2$ .

$f (0) = 5 \sin (\frac{2 (π \cdot (0))}{2 (1)})$

Step 6.1.2.1.2.2

Cancel the common factor.

$f (0) = 5 \sin (\frac{2 (π \cdot (0))}{2 \cdot 1})$

Step 6.1.2.1.2.3

Rewrite the expression.

$f (0) = 5 \sin (\frac{π \cdot (0)}{1})$

Step 6.1.2.1.2.4

Divide $π \cdot (0)$ by $1$ .

$f (0) = 5 \sin (π \cdot (0))$

Step 6.1.2.2

Multiply $π$ by $0$ .

$f (0) = 5 \sin (0)$

Step 6.1.2.3

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

$f (0) = 5 \cdot 0$

Step 6.1.2.4

Multiply $5$ by $0$ .

$f (0) = 0$

Step 6.1.2.5

The final answer is $0$ .

$0$

Step 6.2

Find the point at $x = 1$ .

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Step 6.2.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = 5 \sin (\frac{π (1)}{2})$

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

Multiply $π$ by $1$ .

$f (1) = 5 \sin (\frac{π}{2})$

Step 6.2.2.2

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

$f (1) = 5 \cdot 1$

Step 6.2.2.3

Multiply $5$ by $1$ .

$f (1) = 5$

Step 6.2.2.4

The final answer is $5$ .

$5$

Step 6.3

Find the point at $x = 2$ .

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Step 6.3.1

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

$f (2) = 5 \sin (\frac{π (2)}{2})$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Cancel the common factor of $2$ .

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Step 6.3.2.1.1

Cancel the common factor.

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

Step 6.3.2.1.2

Divide $π$ by $1$ .

$f (2) = 5 \sin (π)$

Step 6.3.2.2

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

$f (2) = 5 \sin (0)$

Step 6.3.2.3

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

$f (2) = 5 \cdot 0$

Step 6.3.2.4

Multiply $5$ by $0$ .

$f (2) = 0$

Step 6.3.2.5

The final answer is $0$ .

$0$

Step 6.4

Find the point at $x = 3$ .

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Step 6.4.1

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

$f (3) = 5 \sin (\frac{π (3)}{2})$

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

Move $3$ to the left of $π$ .

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

Step 6.4.2.2

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.

$f (3) = 5 (- \sin (\frac{π}{2}))$

Step 6.4.2.3

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

$f (3) = 5 (- 1 \cdot 1)$

Step 6.4.2.4

Multiply $5 (- 1 \cdot 1)$ .

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Step 6.4.2.4.1

Multiply $- 1$ by $1$ .

$f (3) = 5 \cdot - 1$

Step 6.4.2.4.2

Multiply $5$ by $- 1$ .

$f (3) = - 5$

Step 6.4.2.5

The final answer is $- 5$ .

$- 5$

Step 6.5

Find the point at $x = 4$ .

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Step 6.5.1

Replace the variable $x$ with $4$ in the expression.

$f (4) = 5 \sin (\frac{π (4)}{2})$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Cancel the common factor of $4$ and $2$ .

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Step 6.5.2.1.1

Factor $2$ out of $π (4)$ .

$f (4) = 5 \sin (\frac{2 (π \cdot (2))}{2})$

Step 6.5.2.1.2

Cancel the common factors.

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Step 6.5.2.1.2.1

Factor $2$ out of $2$ .

$f (4) = 5 \sin (\frac{2 (π \cdot (2))}{2 (1)})$

Step 6.5.2.1.2.2

Cancel the common factor.

$f (4) = 5 \sin (\frac{2 (π \cdot (2))}{2 \cdot 1})$

Step 6.5.2.1.2.3

Rewrite the expression.

$f (4) = 5 \sin (\frac{π \cdot (2)}{1})$

Step 6.5.2.1.2.4

Divide $π \cdot (2)$ by $1$ .

$f (4) = 5 \sin (π \cdot (2))$

Step 6.5.2.2

Move $2$ to the left of $π$ .

$f (4) = 5 \sin (2 \cdot π)$

Step 6.5.2.3

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

$f (4) = 5 \sin (0)$

Step 6.5.2.4

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

$f (4) = 5 \cdot 0$

Step 6.5.2.5

Multiply $5$ by $0$ .

$f (4) = 0$

Step 6.5.2.6

The final answer is $0$ .

$0$

Step 6.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 0 \\ 1 & 5 \\ 2 & 0 \\ 3 & - 5 \\ 4 & 0 \end{array}$

Step 7

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

Amplitude: $5$

Period: $4$

Phase Shift: None

Vertical Shift: None

$\begin{array}{c} x & f (x) \\ 0 & 0 \\ 1 & 5 \\ 2 & 0 \\ 3 & - 5 \\ 4 & 0 \end{array}$

Step 8