Graph y=cos(1/6x)

$y = \cos (\frac{1}{6} x)$

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

$a = 1$

$b = \frac{1}{6}$

$c = 0$

$d = 0$

Find the amplitude $| a |$ .

Amplitude: $1$

Find the period of $\cos (\frac{x}{6})$ .

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

$\frac{2 π}{| b |}$

Replace $b$ with $\frac{1}{6}$ in the formula for period.

$\frac{2 π}{| \frac{1}{6} |}$

$\frac{1}{6}$ is approximately $0.1\overline{6}$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{6}}$

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 6$

Multiply $6$ by $2$ .

$12 π$

Find the phase shift using the formula $\frac{c}{b}$ .

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The phase shift of the function can be calculated from $\frac{c}{b}$ .

Phase Shift: $\frac{c}{b}$

Replace the values of $c$ and $b$ in the equation for phase shift.

Phase Shift: $\frac{0}{\frac{1}{6}}$

Multiply the numerator by the reciprocal of the denominator.

Phase Shift: $0 \cdot 6$

Multiply $0$ by $6$ .

Phase Shift: $0$

Find the vertical shift $d$ .

Vertical Shift: $0$

List the properties of the trigonometric function.

Amplitude: $1$

Period: $12 π$

Phase Shift: $0$ ( $0$ to the right)

Vertical Shift: $0$

Select a few points to graph.

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Find the point at $x = 0$ .

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

$f (0) = \cos (\frac{0}{6})$

Simplify the result.

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Divide $0$ by $6$ .

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

The exact value of $\cos (0)$ is $1$ .

$f (0) = 1$

The final answer is $1$ .

$1$

Find the point at $x = 3 π$ .

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

$f (3 π) = \cos (\frac{3 π}{6})$

Simplify the result.

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Cancel the common factor of $3$ and $6$ .

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Factor $3$ out of $3 π$ .

$f (3 π) = \cos (\frac{3 (π)}{6})$

Cancel the common factors.

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Factor $3$ out of $6$ .

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

Cancel the common factor.

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

Rewrite the expression.

$f (3 π) = \cos (\frac{π}{2})$

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$f (3 π) = 0$

The final answer is $0$ .

$0$

Find the point at $x = 6 π$ .

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Replace the variable $x$ with $6 π$ in the expression.

$f (6 π) = \cos (\frac{6 π}{6})$

Simplify the result.

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Cancel the common factor of $6$ .

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Cancel the common factor.

$f (6 π) = \cos (\frac{6 π}{6})$

Divide $π$ by $1$ .

$f (6 π) = \cos (π)$

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.

$f (6 π) = - \cos (0)$

The exact value of $\cos (0)$ is $1$ .

$f (6 π) = - 1 \cdot 1$

Multiply $- 1$ by $1$ .

$f (6 π) = - 1$

The final answer is $- 1$ .

$- 1$

Find the point at $x = 9 π$ .

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Replace the variable $x$ with $9 π$ in the expression.

$f (9 π) = \cos (\frac{9 π}{6})$

Simplify the result.

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Cancel the common factor of $9$ and $6$ .

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Factor $3$ out of $9 π$ .

$f (9 π) = \cos (\frac{3 (3 π)}{6})$

Cancel the common factors.

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Factor $3$ out of $6$ .

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

Cancel the common factor.

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

Rewrite the expression.

$f (9 π) = \cos (\frac{3 π}{2})$

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

$f (9 π) = \cos (\frac{π}{2})$

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$f (9 π) = 0$

The final answer is $0$ .

$0$

Find the point at $x = 12 π$ .

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Replace the variable $x$ with $12 π$ in the expression.

$f (12 π) = \cos (\frac{12 π}{6})$

Simplify the result.

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Cancel the common factor of $12$ and $6$ .

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Factor $6$ out of $12 π$ .

$f (12 π) = \cos (\frac{6 (2 π)}{6})$

Cancel the common factors.

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Factor $6$ out of $6$ .

$f (12 π) = \cos (\frac{6 (2 π)}{6 (1)})$

Cancel the common factor.

$f (12 π) = \cos (\frac{6 (2 π)}{6 \cdot 1})$

Rewrite the expression.

$f (12 π) = \cos (\frac{2 π}{1})$

Divide $2 π$ by $1$ .

$f (12 π) = \cos (2 π)$

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

$f (12 π) = \cos (0)$

The exact value of $\cos (0)$ is $1$ .

$f (12 π) = 1$

The final answer is $1$ .

$1$

List the points in a table.

$\begin{array}{c} x & f (x) \\ 0 & 1 \\ 3 π & 0 \\ 6 π & - 1 \\ 9 π & 0 \\ 12 π & 1 \end{array}$

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

Amplitude: $1$

Period: $12 π$

Phase Shift: $0$ ( $0$ to the right)

Vertical Shift: $0$

$\begin{array}{c} x & f (x) \\ 0 & 1 \\ 3 π & 0 \\ 6 π & - 1 \\ 9 π & 0 \\ 12 π & 1 \end{array}$

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