Trigonometry Examples

$f (x) = \arccos (2 x)$

Step 1

Select a few points to graph.

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Step 1.1

Find the point at $x = - \frac{1}{2}$ .

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Step 1.1.1

Replace the variable $x$ with $- \frac{1}{2}$ in the expression.

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

Step 1.1.2

Simplify the result.

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Step 1.1.2.1

Cancel the common factor of $2$ .

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Step 1.1.2.1.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Step 1.1.2.1.2

Cancel the common factor.

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

Step 1.1.2.1.3

Rewrite the expression.

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

Step 1.1.2.2

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

$f (- \frac{1}{2}) = π$

Step 1.1.2.3

The final answer is $π$ .

$π$

Step 1.2

Find the point at $x = - \frac{1}{4}$ .

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Step 1.2.1

Replace the variable $x$ with $- \frac{1}{4}$ in the expression.

$f (- \frac{1}{4}) = \arccos (2 (- \frac{1}{4}))$

Step 1.2.2

Simplify the result.

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Step 1.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.2.1.1

Move the leading negative in $- \frac{1}{4}$ into the numerator.

$f (- \frac{1}{4}) = \arccos (2 (\frac{- 1}{4}))$

Step 1.2.2.1.2

Factor $2$ out of $4$ .

$f (- \frac{1}{4}) = \arccos (2 (\frac{- 1}{2 (2)}))$

Step 1.2.2.1.3

Cancel the common factor.

$f (- \frac{1}{4}) = \arccos (2 (\frac{- 1}{2 \cdot 2}))$

Step 1.2.2.1.4

Rewrite the expression.

$f (- \frac{1}{4}) = \arccos (\frac{- 1}{2})$

Step 1.2.2.2

Move the negative in front of the fraction.

$f (- \frac{1}{4}) = \arccos (- \frac{1}{2})$

Step 1.2.2.3

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

$f (- \frac{1}{4}) = \frac{2 π}{3}$

Step 1.2.2.4

The final answer is $\frac{2 π}{3}$ .

$\frac{2 π}{3}$

Step 1.3

Find the point at $x = 0$ .

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Step 1.3.1

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

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

Step 1.3.2

Simplify the result.

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Step 1.3.2.1

Multiply $2$ by $0$ .

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

Step 1.3.2.2

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

$f (0) = \frac{π}{2}$

Step 1.3.2.3

The final answer is $\frac{π}{2}$ .

$\frac{π}{2}$

Step 1.4

Find the point at $x = \frac{1}{4}$ .

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Step 1.4.1

Replace the variable $x$ with $\frac{1}{4}$ in the expression.

$f (\frac{1}{4}) = \arccos (2 (\frac{1}{4}))$

Step 1.4.2

Simplify the result.

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Step 1.4.2.1

Cancel the common factor of $2$ .

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Step 1.4.2.1.1

Factor $2$ out of $4$ .

$f (\frac{1}{4}) = \arccos (2 (\frac{1}{2 (2)}))$

Step 1.4.2.1.2

Cancel the common factor.

$f (\frac{1}{4}) = \arccos (2 (\frac{1}{2 \cdot 2}))$

Step 1.4.2.1.3

Rewrite the expression.

$f (\frac{1}{4}) = \arccos (\frac{1}{2})$

Step 1.4.2.2

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

$f (\frac{1}{4}) = \frac{π}{3}$

Step 1.4.2.3

The final answer is $\frac{π}{3}$ .

$\frac{π}{3}$

Step 1.5

Find the point at $x = \frac{1}{2}$ .

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Step 1.5.1

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

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

Step 1.5.2

Simplify the result.

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Step 1.5.2.1

Cancel the common factor of $2$ .

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Step 1.5.2.1.1

Cancel the common factor.

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

Step 1.5.2.1.2

Rewrite the expression.

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

Step 1.5.2.2

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

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

Step 1.5.2.3

The final answer is $0$ .

$0$

Step 1.6

List the points in a table.

$\begin{array}{c} x & f (x) \\ - \frac{1}{2} & π \\ - \frac{1}{4} & \frac{2 π}{3} \\ 0 & \frac{π}{2} \\ \frac{1}{4} & \frac{π}{3} \\ \frac{1}{2} & 0 \end{array}$

Step 2

The trig function can be graphed using the points.

$\begin{array}{c} x & f (x) \\ - \frac{1}{2} & π \\ - \frac{1}{4} & \frac{2 π}{3} \\ 0 & \frac{π}{2} \\ \frac{1}{4} & \frac{π}{3} \\ \frac{1}{2} & 0 \end{array}$

Step 3