Trigonometry Examples

$y = \cos (π \cdot x)$

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.

$a = 1$

$b = π$

$c = 0$

$d = 0$

Step 2

Find the amplitude

$| a |$ .

Amplitude:

$1$

Step 3

Find the period of

$\cos (π \cdot x)$ .

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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

$π$ in the formula for period.

$\frac{2 π}{| π |}$

Step 3.3

$π$ is approximately

$3.14159265$ which is positive so remove the absolute value

$\frac{2 π}{π}$

Step 3.4

Cancel the common factor of

$π$ .

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Step 3.4.1

Cancel the common factor.

$\frac{2 π}{π}$

Step 3.4.2

Divide

$2$ by

$1$ .

$2$

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:

$\frac{0}{π}$

Step 4.3

Divide

$0$ by

$π$ .

Phase Shift:

$0$

Phase Shift:

$0$

Step 5

List the properties of the trigonometric function.

Amplitude:

$1$

Period:

$2$

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.

$f (0) = \cos (π \cdot (0))$

Step 6.1.2

Simplify the result.

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Step 6.1.2.1

Multiply

$π$ by

$0$ .

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

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

$x = \frac{1}{2}$ .

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Step 6.2.1

Replace the variable

$x$ with

$\frac{1}{2}$ in the expression.

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

Step 6.2.2

Simplify the result.

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Step 6.2.2.1

Combine

$π$ and

$\frac{1}{2}$ .

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

Step 6.2.2.2

The exact value of

$\cos (\frac{π}{2})$ is

$0$ .

$f (\frac{1}{2}) = 0$

Step 6.2.2.3

The final answer is

$0$ .

$0$

Step 6.3

Find the point at

$x = 1$ .

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Step 6.3.1

Replace the variable

$x$ with

$1$ in the expression.

$f (1) = \cos (π \cdot (1))$

Step 6.3.2

Simplify the result.

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Step 6.3.2.1

Multiply

$π$ by

$1$ .

$f (1) = \cos (π)$

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.

$f (1) = - \cos (0)$

Step 6.3.2.3

The exact value of

$\cos (0)$ is

$1$ .

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

Step 6.3.2.4

Multiply

$- 1$ by

$1$ .

$f (1) = - 1$

Step 6.3.2.5

The final answer is

$- 1$ .

$- 1$

Step 6.4

Find the point at

$x = \frac{3}{2}$ .

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Step 6.4.1

Replace the variable

$x$ with

$\frac{3}{2}$ in the expression.

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

Step 6.4.2

Simplify the result.

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Step 6.4.2.1

Combine

$π$ and

$\frac{3}{2}$ .

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

Step 6.4.2.2

Move

$3$ to the left of

$π$ .

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

Step 6.4.2.3

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

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

Step 6.4.2.4

The exact value of

$\cos (\frac{π}{2})$ is

$0$ .

$f (\frac{3}{2}) = 0$

Step 6.4.2.5

The final answer is

$0$ .

$0$

Step 6.5

Find the point at

$x = 2$ .

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Step 6.5.1

Replace the variable

$x$ with

$2$ in the expression.

$f (2) = \cos (π \cdot (2))$

Step 6.5.2

Simplify the result.

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Step 6.5.2.1

Move

$2$ to the left of

$π$ .

$f (2) = \cos (2 \cdot π)$

Step 6.5.2.2

Subtract full rotations of

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

$0$ and less than

$2 π$ .

$f (2) = \cos (0)$

Step 6.5.2.3

The exact value of

$\cos (0)$ is

$1$ .

$f (2) = 1$

Step 6.5.2.4

The final answer is

$1$ .

$1$

Step 6.6

List the points in a table.

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

Step 7

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

Amplitude:

$1$

Period:

$2$

Phase Shift: None

Vertical Shift: None

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

Step 8