Trigonometry Examples

$y = \cos (\frac{1}{3} x)$

Step 1

Use the form

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

$a = 1$

$b = \frac{1}{3}$

$c = 0$

$d = 0$

Step 2

Find the amplitude

$| a |$ .

Amplitude:

$1$

Step 3

Find the period of

$\cos (\frac{x}{3})$ .

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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

$\frac{1}{3}$ in the formula for period.

$\frac{2 π}{| \frac{1}{3} |}$

Step 3.3

$\frac{1}{3}$ is approximately

$0.\overline{3}$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{3}}$

Step 3.4

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 3$

Step 3.5

Multiply

$3$ by

$2$ .

$6 π$

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:

$\frac{0}{\frac{1}{3}}$

Step 4.3

Multiply the numerator by the reciprocal of the denominator.

Phase Shift:

$0 \cdot 3$

Step 4.4

Multiply

$0$ by

$3$ .

Phase Shift:

$0$

Phase Shift:

$0$

Step 5

List the properties of the trigonometric function.

Amplitude:

$1$

Period:

$6 π$

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.

$f (0) = \cos (\frac{0}{3})$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Divide

$0$ by

$3$ .

$f (0) = \cos (0)$

Step 6.1.2.2

The exact value of

$\cos (0)$ is

$1$ .

$f (0) = 1$

Step 6.1.2.3

The final answer is

$1$ .

$1$

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.

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

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 6.2.2.2

Cancel the common factor of

$3$ .

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Step 6.2.2.2.1

Factor

$3$ out of

$3 π$ .

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

Step 6.2.2.2.2

Cancel the common factor.

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

Step 6.2.2.2.3

Rewrite the expression.

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

Step 6.2.2.3

The exact value of

$\cos (\frac{π}{2})$ is

$0$ .

$f (\frac{3 π}{2}) = 0$

Step 6.2.2.4

The final answer is

$0$ .

$0$

Step 6.3

Find the point at

$x = 3 π$ .

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Step 6.3.1

Replace the variable

$x$ with

$3 π$ in the expression.

$f (3 π) = \cos (\frac{3 π}{3})$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Cancel the common factor of

$3$ .

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Step 6.3.2.1.1

Cancel the common factor.

$f (3 π) = \cos (\frac{3 π}{3})$

Step 6.3.2.1.2

Divide

$π$ by

$1$ .

$f (3 π) = \cos (π)$

Step 6.3.2.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$f (3 π) = - \cos (0)$

Step 6.3.2.3

The exact value of

$\cos (0)$ is

$1$ .

$f (3 π) = - 1 \cdot 1$

Step 6.3.2.4

Multiply

$- 1$ by

$1$ .

$f (3 π) = - 1$

Step 6.3.2.5

The final answer is

$- 1$ .

$- 1$

Step 6.4

Find the point at

$x = \frac{9 π}{2}$ .

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Step 6.4.1

Replace the variable

$x$ with

$\frac{9 π}{2}$ in the expression.

$f (\frac{9 π}{2}) = \cos (\frac{\frac{9 π}{2}}{3})$

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{9 π}{2}) = \cos (\frac{9 π}{2} \cdot \frac{1}{3})$

Step 6.4.2.2

Cancel the common factor of

$3$ .

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Step 6.4.2.2.1

Factor

$3$ out of

$9 π$ .

$f (\frac{9 π}{2}) = \cos (\frac{3 (3 π)}{2} \cdot \frac{1}{3})$

Step 6.4.2.2.2

Cancel the common factor.

$f (\frac{9 π}{2}) = \cos (\frac{3 (3 π)}{2} \cdot \frac{1}{3})$

Step 6.4.2.2.3

Rewrite the expression.

$f (\frac{9 π}{2}) = \cos (\frac{3 π}{2})$

Step 6.4.2.3

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

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

Step 6.4.2.4

The exact value of

$\cos (\frac{π}{2})$ is

$0$ .

$f (\frac{9 π}{2}) = 0$

Step 6.4.2.5

The final answer is

$0$ .

$0$

Step 6.5

Find the point at

$x = 6 π$ .

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Step 6.5.1

Replace the variable

$x$ with

$6 π$ in the expression.

$f (6 π) = \cos (\frac{6 π}{3})$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Cancel the common factor of

$6$ and

$3$ .

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Step 6.5.2.1.1

Factor

$3$ out of

$6 π$ .

$f (6 π) = \cos (\frac{3 (2 π)}{3})$

Step 6.5.2.1.2

Cancel the common factors.

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Step 6.5.2.1.2.1

Factor

$3$ out of

$3$ .

$f (6 π) = \cos (\frac{3 (2 π)}{3 (1)})$

Step 6.5.2.1.2.2

Cancel the common factor.

$f (6 π) = \cos (\frac{3 (2 π)}{3 \cdot 1})$

Step 6.5.2.1.2.3

Rewrite the expression.

$f (6 π) = \cos (\frac{2 π}{1})$

Step 6.5.2.1.2.4

Divide

$2 π$ by

$1$ .

$f (6 π) = \cos (2 π)$

Step 6.5.2.2

Subtract full rotations of

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

$0$ and less than

$2 π$ .

$f (6 π) = \cos (0)$

Step 6.5.2.3

The exact value of

$\cos (0)$ is

$1$ .

$f (6 π) = 1$

Step 6.5.2.4

The final answer is

$1$ .

$1$

Step 6.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 1 \\ \frac{3 π}{2} & 0 \\ 3 π & - 1 \\ \frac{9 π}{2} & 0 \\ 6 π & 1 \end{array}$

Step 7

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

Amplitude:

$1$

Period:

$6 π$

Phase Shift: None

Vertical Shift: None

$\begin{array}{c} x & f (x) \\ 0 & 1 \\ \frac{3 π}{2} & 0 \\ 3 π & - 1 \\ \frac{9 π}{2} & 0 \\ 6 π & 1 \end{array}$

Step 8