Trigonometry Examples

$\arccos (x + 2)$

Step 1

Select a few points to graph.

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Step 1.1

Find the point at $x = - 3$ .

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Step 1.1.1

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

$f (- 3) = \arccos ((- 3) + 2)$

Step 1.1.2

Simplify the result.

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Step 1.1.2.1

Add $- 3$ and $2$ .

$f (- 3) = \arccos (- 1)$

Step 1.1.2.2

The exact value of $\arccos (- 1)$ is $π$ .

$f (- 3) = π$

Step 1.1.2.3

The final answer is $π$ .

$π$

Step 1.2

Find the point at $x = - \frac{5}{2}$ .

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Step 1.2.1

Replace the variable $x$ with $- \frac{5}{2}$ in the expression.

$f (- \frac{5}{2}) = \arccos ((- \frac{5}{2}) + 2)$

Step 1.2.2

Simplify the result.

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Step 1.2.2.1

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (- \frac{5}{2}) = \arccos (- \frac{5}{2} + 2 \cdot \frac{2}{2})$

Step 1.2.2.2

Combine $2$ and $\frac{2}{2}$ .

$f (- \frac{5}{2}) = \arccos (- \frac{5}{2} + \frac{2 \cdot 2}{2})$

Step 1.2.2.3

Combine the numerators over the common denominator.

$f (- \frac{5}{2}) = \arccos (\frac{- 5 + 2 \cdot 2}{2})$

Step 1.2.2.4

Simplify the numerator.

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Step 1.2.2.4.1

Multiply $2$ by $2$ .

$f (- \frac{5}{2}) = \arccos (\frac{- 5 + 4}{2})$

Step 1.2.2.4.2

Add $- 5$ and $4$ .

$f (- \frac{5}{2}) = \arccos (\frac{- 1}{2})$

Step 1.2.2.5

Move the negative in front of the fraction.

$f (- \frac{5}{2}) = \arccos (- \frac{1}{2})$

Step 1.2.2.6

The exact value of $\arccos (- \frac{1}{2})$ is $\frac{2 π}{3}$ .

$f (- \frac{5}{2}) = \frac{2 π}{3}$

Step 1.2.2.7

The final answer is $\frac{2 π}{3}$ .

$\frac{2 π}{3}$

Step 1.3

Find the point at $x = - 2$ .

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Step 1.3.1

Replace the variable $x$ with $- 2$ in the expression.

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

Step 1.3.2

Simplify the result.

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Step 1.3.2.1

Add $- 2$ and $2$ .

$f (- 2) = \arccos (0)$

Step 1.3.2.2

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

$f (- 2) = \frac{π}{2}$

Step 1.3.2.3

The final answer is $\frac{π}{2}$ .

$\frac{π}{2}$

Step 1.4

Find the point at $x = - \frac{3}{2}$ .

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Step 1.4.1

Replace the variable $x$ with $- \frac{3}{2}$ in the expression.

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

Step 1.4.2

Simplify the result.

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Step 1.4.2.1

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 1.4.2.2

Combine $2$ and $\frac{2}{2}$ .

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

Step 1.4.2.3

Combine the numerators over the common denominator.

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

Step 1.4.2.4

Simplify the numerator.

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Step 1.4.2.4.1

Multiply $2$ by $2$ .

$f (- \frac{3}{2}) = \arccos (\frac{- 3 + 4}{2})$

Step 1.4.2.4.2

Add $- 3$ and $4$ .

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

Step 1.4.2.5

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$f (- \frac{3}{2}) = \frac{π}{3}$

Step 1.4.2.6

The final answer is $\frac{π}{3}$ .

$\frac{π}{3}$

Step 1.5

Find the point at $x = - 1$ .

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Step 1.5.1

Replace the variable $x$ with $- 1$ in the expression.

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

Step 1.5.2

Simplify the result.

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Step 1.5.2.1

Add $- 1$ and $2$ .

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

Step 1.5.2.2

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

$f (- 1) = 0$

Step 1.5.2.3

The final answer is $0$ .

$0$

Step 1.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ - 3 & π \\ - \frac{5}{2} & \frac{2 π}{3} \\ - 2 & \frac{π}{2} \\ - \frac{3}{2} & \frac{π}{3} \\ - 1 & 0 \end{array}$

Step 2

The trig function can be graphed using the points.

$\begin{array}{c} x & f (x) \\ - 3 & π \\ - \frac{5}{2} & \frac{2 π}{3} \\ - 2 & \frac{π}{2} \\ - \frac{3}{2} & \frac{π}{3} \\ - 1 & 0 \end{array}$

Step 3