Trigonometry Examples

y = cos (\frac{π}{18} - \frac{x}{3}) + 2

$y = \cos (\frac{π}{18} - \frac{x}{3}) + 2$

Step 1

Use the form

a cos (b x - c) + d

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

a = 1

$a = 1$

b = - \frac{1}{3}

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

c = - \frac{π}{18}

$c = - \frac{π}{18}$

d = 2

$d = 2$

Step 2

Find the amplitude

| a |

$| a |$ .

Amplitude:

1

$1$

Step 3

Find the period using the formula

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

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Step 3.1

Find the period of

cos (\frac{π}{18} - \frac{x}{3})

$\cos (\frac{π}{18} - \frac{x}{3})$ .

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Step 3.1.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 3.1.2

Replace

b

$b$ with

- \frac{1}{3}

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

\frac{2 π}{∣ ∣ - \frac{1}{3} ∣ ∣}

$\frac{2 π}{| - \frac{1}{3} |}$

Step 3.1.3

- \frac{1}{3}

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

- 0. ¯ 3

$- 0.\overline{3}$ which is negative so negate

- \frac{1}{3}

$- \frac{1}{3}$ and remove the absolute value

\frac{2 π}{\frac{1}{3}}

$\frac{2 π}{\frac{1}{3}}$

Step 3.1.4

Multiply the numerator by the reciprocal of the denominator.

2 π \cdot 3

$2 π \cdot 3$

Step 3.1.5

Multiply

3

$3$ by

2

$2$ .

6 π

$6 π$

6 π

$6 π$

Step 3.2

Find the period of

2

$2$ .

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Step 3.2.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 3.2.2

Replace

b

$b$ with

- \frac{1}{3}

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

\frac{2 π}{∣ ∣ - \frac{1}{3} ∣ ∣}

$\frac{2 π}{| - \frac{1}{3} |}$

Step 3.2.3

- \frac{1}{3}

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

- 0. ¯ 3

$- 0.\overline{3}$ which is negative so negate

- \frac{1}{3}

$- \frac{1}{3}$ and remove the absolute value

\frac{2 π}{\frac{1}{3}}

$\frac{2 π}{\frac{1}{3}}$

Step 3.2.4

Multiply the numerator by the reciprocal of the denominator.

2 π \cdot 3

$2 π \cdot 3$

Step 3.2.5

Multiply

3

$3$ by

2

$2$ .

6 π

$6 π$

6 π

$6 π$

Step 3.3

The period of addition/subtraction of trig functions is the maximum of the individual periods.

6 π

$6 π$

6 π

$6 π$

Step 4

Find the phase shift using the formula

\frac{c}{b}

$\frac{c}{b}$ .

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Step 4.1

The phase shift of the function can be calculated from

\frac{c}{b}

$\frac{c}{b}$ .

Phase Shift:

\frac{c}{b}

$\frac{c}{b}$

Step 4.2

Replace the values of

c

$c$ and

b

$b$ in the equation for phase shift.

Phase Shift:

\frac{- \frac{π}{18}}{- \frac{1}{3}}

$\frac{- \frac{π}{18}}{- \frac{1}{3}}$

Step 4.3

Dividing two negative values results in a positive value.

Phase Shift:

\frac{\frac{π}{18}}{\frac{1}{3}}

$\frac{\frac{π}{18}}{\frac{1}{3}}$

Step 4.4

Multiply the numerator by the reciprocal of the denominator.

Phase Shift:

\frac{π}{18} \cdot 3

$\frac{π}{18} \cdot 3$

Step 4.5

Cancel the common factor of

3

$3$ .

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Step 4.5.1

Factor

3

$3$ out of

18

$18$ .

Phase Shift:

\frac{π}{3 (6)} \cdot 3

$\frac{π}{3 (6)} \cdot 3$

Step 4.5.2

Cancel the common factor.

Phase Shift:

$\frac{π}{3 \cdot 6} \cdot 3$

Step 4.5.3

Rewrite the expression.

Phase Shift:

$\frac{π}{6}$

Phase Shift:

$\frac{π}{6}$

Phase Shift:

$\frac{π}{6}$

Step 5

List the properties of the trigonometric function.

Amplitude:

$1$

Period:

$6 π$

Phase Shift:

$\frac{π}{6}$ (

$\frac{π}{6}$ to the right)

Vertical Shift:

$2$

Step 6