Trigonometry Examples

y = sin (9 x)

$y = \sin (9 x)$

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.

$a = 1$

$b = 9$

$c = 0$

$d = 0$

Step 2

Find the amplitude

$| a |$ .

Amplitude:

$1$

Step 3

Find the period of

$\sin (9 x)$ .

Tap for more steps...

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

$9$ in the formula for period.

$\frac{2 π}{| 9 |}$

Step 3.3

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

$0$ and

$9$ is

$9$ .

$\frac{2 π}{9}$

Step 4

Find the phase shift using the formula

$\frac{c}{b}$ .

Tap for more steps...

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:

$\frac{0}{9}$

Step 4.3

Divide

$0$ by

$9$ .

Phase Shift:

$0$

Phase Shift:

$0$

Step 5

List the properties of the trigonometric function.

Amplitude:

$1$

Period:

$\frac{2 π}{9}$

Phase Shift: None

Vertical Shift: None

Step 6

Select a few points to graph.

Tap for more steps...

Step 6.1

Find the point at

$x = 0$ .

Tap for more steps...

Step 6.1.1

Replace the variable

$x$ with

$0$ in the expression.

$f (0) = \sin (9 (0))$

Step 6.1.2

Simplify the result.

Tap for more steps...

Step 6.1.2.1

Multiply

$9$ by

$0$ .

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

Step 6.1.2.2

The exact value of

$\sin (0)$ is

$0$ .

$f (0) = 0$

Step 6.1.2.3

The final answer is

$0$ .

$0$

Step 6.2

Find the point at

$x = \frac{π}{18}$ .

Tap for more steps...

Step 6.2.1

Replace the variable

$x$ with

$\frac{π}{18}$ in the expression.

$f (\frac{π}{18}) = \sin (9 (\frac{π}{18}))$

Step 6.2.2

Simplify the result.

Tap for more steps...

Step 6.2.2.1

Cancel the common factor of

$9$ .

Tap for more steps...

Step 6.2.2.1.1

Factor

$9$ out of

$18$ .

$f (\frac{π}{18}) = \sin (9 (\frac{π}{9 (2)}))$

Step 6.2.2.1.2

Cancel the common factor.

$f (\frac{π}{18}) = \sin (9 (\frac{π}{9 \cdot 2}))$

Step 6.2.2.1.3

Rewrite the expression.

$f (\frac{π}{18}) = \sin (\frac{π}{2})$

Step 6.2.2.2

The exact value of

$\sin (\frac{π}{2})$ is

$1$ .

$f (\frac{π}{18}) = 1$

Step 6.2.2.3

The final answer is

$1$ .

$1$

Step 6.3

Find the point at

$x = \frac{π}{9}$ .

Tap for more steps...

Step 6.3.1

Replace the variable

$x$ with

$\frac{π}{9}$ in the expression.

$f (\frac{π}{9}) = \sin (9 (\frac{π}{9}))$

Step 6.3.2

Simplify the result.

Tap for more steps...

Step 6.3.2.1

Cancel the common factor of

$9$ .

Tap for more steps...

Step 6.3.2.1.1

Cancel the common factor.

$f (\frac{π}{9}) = \sin (9 (\frac{π}{9}))$

Step 6.3.2.1.2

Rewrite the expression.

$f (\frac{π}{9}) = \sin (π)$

Step 6.3.2.2

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

$f (\frac{π}{9}) = \sin (0)$

Step 6.3.2.3

The exact value of

$\sin (0)$ is

$0$ .

$f (\frac{π}{9}) = 0$

Step 6.3.2.4

The final answer is

$0$ .

$0$

Step 6.4

Find the point at

$x = \frac{π}{6}$ .

Tap for more steps...

Step 6.4.1

Replace the variable

$x$ with

$\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = \sin (9 (\frac{π}{6}))$

Step 6.4.2

Simplify the result.

Tap for more steps...

Step 6.4.2.1

Cancel the common factor of

$3$ .

Tap for more steps...

Step 6.4.2.1.1

Factor

$3$ out of

$9$ .

$f (\frac{π}{6}) = \sin (3 (3) (\frac{π}{6}))$

Step 6.4.2.1.2

Factor

$3$ out of

$6$ .

$f (\frac{π}{6}) = \sin (3 \cdot (3 (\frac{π}{3 \cdot 2})))$

Step 6.4.2.1.3

Cancel the common factor.

$f (\frac{π}{6}) = \sin (3 \cdot (3 (\frac{π}{3 \cdot 2})))$

Step 6.4.2.1.4

Rewrite the expression.

$f (\frac{π}{6}) = \sin (3 (\frac{π}{2}))$

Step 6.4.2.2

Combine

$3$ and

$\frac{π}{2}$ .

$f (\frac{π}{6}) = \sin (\frac{3 π}{2})$

Step 6.4.2.3

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.

$f (\frac{π}{6}) = - \sin (\frac{π}{2})$

Step 6.4.2.4

The exact value of

$\sin (\frac{π}{2})$ is

$1$ .

$f (\frac{π}{6}) = - 1 \cdot 1$

Step 6.4.2.5

Multiply

$- 1$ by

$1$ .

$f (\frac{π}{6}) = - 1$

Step 6.4.2.6

The final answer is

$- 1$ .

$- 1$

Step 6.5

Find the point at

$x = \frac{2 π}{9}$ .

Tap for more steps...

Step 6.5.1

Replace the variable

$x$ with

$\frac{2 π}{9}$ in the expression.

$f (\frac{2 π}{9}) = \sin (9 (\frac{2 π}{9}))$

Step 6.5.2

Simplify the result.

Tap for more steps...

Step 6.5.2.1

Cancel the common factor of

$9$ .

Tap for more steps...

Step 6.5.2.1.1

Cancel the common factor.

$f (\frac{2 π}{9}) = \sin (9 (\frac{2 π}{9}))$

Step 6.5.2.1.2

Rewrite the expression.

$f (\frac{2 π}{9}) = \sin (2 π)$

Step 6.5.2.2

Subtract full rotations of

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

$0$ and less than

$2 π$ .

$f (\frac{2 π}{9}) = \sin (0)$

Step 6.5.2.3

The exact value of

$\sin (0)$ is

$0$ .

$f (\frac{2 π}{9}) = 0$

Step 6.5.2.4

The final answer is

$0$ .

$0$

Step 6.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 0 \\ \frac{π}{18} & 1 \\ \frac{π}{9} & 0 \\ \frac{π}{6} & - 1 \\ \frac{2 π}{9} & 0 \end{array}$

Step 7

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

Amplitude:

$1$

Period:

$\frac{2 π}{9}$

Phase Shift: None

Vertical Shift: None

$\begin{array}{c} x & f (x) \\ 0 & 0 \\ \frac{π}{18} & 1 \\ \frac{π}{9} & 0 \\ \frac{π}{6} & - 1 \\ \frac{2 π}{9} & 0 \end{array}$

Step 8