Trigonometry Examples

f (x) = arccos (x + 1)

$f (x) = \arccos (x + 1)$

Step 1

Select a few points to graph.

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Step 1.1

Find the point at

x = - 2

$x = - 2$ .

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Step 1.1.1

Replace the variable

x

$x$ with

- 2

$- 2$ in the expression.

f (- 2) = arccos ((- 2) + 1)

$f (- 2) = \arccos ((- 2) + 1)$

Step 1.1.2

Simplify the result.

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Step 1.1.2.1

Add

- 2

$- 2$ and

1

$1$ .

f (- 2) = arccos (- 1)

$f (- 2) = \arccos (- 1)$

Step 1.1.2.2

The exact value of

arccos (- 1)

$\arccos (- 1)$ is

π

$π$ .

f (- 2) = π

$f (- 2) = π$

Step 1.1.2.3

The final answer is

π

$π$ .

π

$π$

π

$π$

π

$π$

Step 1.2

Find the point at

x = - \frac{3}{2}

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

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Step 1.2.1

Replace the variable

x

$x$ with

- \frac{3}{2}

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

f (- \frac{3}{2}) = arccos ((- \frac{3}{2}) + 1)

$f (- \frac{3}{2}) = \arccos ((- \frac{3}{2}) + 1)$

Step 1.2.2

Simplify the result.

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Step 1.2.2.1

Write

1

$1$ as a fraction with a common denominator.

f (- \frac{3}{2}) = arccos (- \frac{3}{2} + \frac{2}{2})

$f (- \frac{3}{2}) = \arccos (- \frac{3}{2} + \frac{2}{2})$

Step 1.2.2.2

Combine the numerators over the common denominator.

f (- \frac{3}{2}) = arccos (\frac{- 3 + 2}{2})

$f (- \frac{3}{2}) = \arccos (\frac{- 3 + 2}{2})$

Step 1.2.2.3

Add

- 3

$- 3$ and

2

$2$ .

f (- \frac{3}{2}) = arccos (\frac{- 1}{2})

$f (- \frac{3}{2}) = \arccos (\frac{- 1}{2})$

Step 1.2.2.4

Move the negative in front of the fraction.

f (- \frac{3}{2}) = arccos (- \frac{1}{2})

$f (- \frac{3}{2}) = \arccos (- \frac{1}{2})$

Step 1.2.2.5

The exact value of

arccos (- \frac{1}{2})

$\arccos (- \frac{1}{2})$ is

\frac{2 π}{3}

$\frac{2 π}{3}$ .

f (- \frac{3}{2}) = \frac{2 π}{3}

$f (- \frac{3}{2}) = \frac{2 π}{3}$

Step 1.2.2.6

The final answer is

\frac{2 π}{3}

$\frac{2 π}{3}$ .

\frac{2 π}{3}

$\frac{2 π}{3}$

\frac{2 π}{3}

$\frac{2 π}{3}$

\frac{2 π}{3}

$\frac{2 π}{3}$

Step 1.3

Find the point at

x = - 1

$x = - 1$ .

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Step 1.3.1

Replace the variable

x

$x$ with

- 1

$- 1$ in the expression.

f (- 1) = arccos ((- 1) + 1)

$f (- 1) = \arccos ((- 1) + 1)$

Step 1.3.2

Simplify the result.

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Step 1.3.2.1

Add

- 1

$- 1$ and

1

$1$ .

f (- 1) = arccos (0)

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

Step 1.3.2.2

The exact value of

arccos (0)

$\arccos (0)$ is

\frac{π}{2}

$\frac{π}{2}$ .

f (- 1) = \frac{π}{2}

$f (- 1) = \frac{π}{2}$

Step 1.3.2.3

The final answer is

$\frac{π}{2}$ .

$\frac{π}{2}$

Step 1.4

Find the point at

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

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Step 1.4.1

Replace the variable

$x$ with

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

$f (- \frac{1}{2}) = \arccos ((- \frac{1}{2}) + 1)$

Step 1.4.2

Simplify the result.

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Step 1.4.2.1

Write

$1$ as a fraction with a common denominator.

$f (- \frac{1}{2}) = \arccos (- \frac{1}{2} + \frac{2}{2})$

Step 1.4.2.2

Combine the numerators over the common denominator.

$f (- \frac{1}{2}) = \arccos (\frac{- 1 + 2}{2})$

Step 1.4.2.3

Add

$- 1$ and

$2$ .

$f (- \frac{1}{2}) = \arccos (\frac{1}{2})$

Step 1.4.2.4

The exact value of

$\arccos (\frac{1}{2})$ is

$\frac{π}{3}$ .

$f (- \frac{1}{2}) = \frac{π}{3}$

Step 1.4.2.5

The final answer is

$\frac{π}{3}$ .

$\frac{π}{3}$

Step 1.5

Find the point at

$x = 0$ .

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Step 1.5.1

Replace the variable

$x$ with

$0$ in the expression.

$f (0) = \arccos ((0) + 1)$

Step 1.5.2

Simplify the result.

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Step 1.5.2.1

Add

$0$ and

$1$ .

$f (0) = \arccos (1)$

Step 1.5.2.2

The exact value of

$\arccos (1)$ is

$0$ .

$f (0) = 0$

Step 1.5.2.3

The final answer is

$0$ .

$0$

Step 1.6

List the points in a table.

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

Step 2

The trig function can be graphed using the points.

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

Step 3