Trigonometry Examples

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

Step 1

Write $f (x) = \frac{1}{3} \cdot \arccos (x - 2)$ as an equation.

$y = \frac{1}{3} \cdot \arccos (x - 2)$

Step 2

Interchange the variables.

$x = \frac{1}{3} \cdot \arccos (y - 2)$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{1}{3} \cdot \arccos (y - 2) = x$ .

$\frac{1}{3} \cdot \arccos (y - 2) = x$

Step 3.2

Multiply each term in $\frac{1}{3} \cdot \arccos (y - 2) = x$ by $3$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $\frac{1}{3} \cdot \arccos (y - 2) = x$ by $3$ .

$\frac{1}{3} \cdot \arccos (y - 2) \cdot 3 = x \cdot 3$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Combine $\frac{1}{3}$ and $\arccos (y - 2)$ .

$\frac{\arccos (y - 2)}{3} \cdot 3 = x \cdot 3$

Step 3.2.2.2

Cancel the common factor of $3$ .

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Step 3.2.2.2.1

Cancel the common factor.

$\frac{\arccos (y - 2)}{3} \cdot 3 = x \cdot 3$

Step 3.2.2.2.2

Rewrite the expression.

$\arccos (y - 2) = x \cdot 3$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Move $3$ to the left of $x$ .

$\arccos (y - 2) = 3 x$

Step 3.3

Take the inverse arccosine of both sides of the equation to extract $y$ from inside the arccosine.

$y - 2 = \cos (3 x)$

Step 3.4

Add $2$ to both sides of the equation.

$y = \cos (3 x) + 2$

Step 4

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \cos (3 x) + 2$

Step 5

Verify if $f^{- 1} (x) = \cos (3 x) + 2$ is the inverse of $f (x) = \frac{1}{3} \cdot \arccos (x - 2)$ .

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Step 5.1

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Step 5.2

Evaluate $f^{- 1} (f (x))$ .

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Step 5.2.1

Set up the composite result function.

$f^{- 1} (f (x))$

Step 5.2.2

Evaluate $f^{- 1} (\frac{1}{3} \cdot \arccos (x - 2))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{1}{3} \cdot \arccos (x - 2)) = \cos (3 (\frac{1}{3} \cdot \arccos (x - 2))) + 2$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Combine $\frac{1}{3}$ and $\arccos (x - 2)$ .

$f^{- 1} (\frac{1}{3} \cdot \arccos (x - 2)) = \cos (3 (\frac{\arccos (x - 2)}{3})) + 2$

Step 5.2.3.2

Cancel the common factor of $3$ .

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Step 5.2.3.2.1

Cancel the common factor.

$f^{- 1} (\frac{1}{3} \cdot \arccos (x - 2)) = \cos (3 (\frac{\arccos (x - 2)}{3})) + 2$

Step 5.2.3.2.2

Rewrite the expression.

$f^{- 1} (\frac{1}{3} \cdot \arccos (x - 2)) = \cos (\arccos (x - 2)) + 2$

Step 5.2.3.3

The functions cosine and arccosine are inverses.

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

Step 5.2.4

Combine the opposite terms in $x - 2 + 2$ .

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Step 5.2.4.1

Add $- 2$ and $2$ .

$f^{- 1} (\frac{1}{3} \cdot \arccos (x - 2)) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

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

Step 5.3

Evaluate $f (f^{- 1} (x))$ .

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Step 5.3.1

Set up the composite result function.

$f (f^{- 1} (x))$

Step 5.3.2

Evaluate $f (\cos (3 x) + 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Step 5.3.3

Combine the opposite terms in $(\cos (3 x) + 2) - 2$ .

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Step 5.3.3.1

Subtract $2$ from $2$ .

$f (\cos (3 x) + 2) = \frac{1}{3} \cdot \arccos (\cos (3 x) + 0)$

Step 5.3.3.2

Add $\cos (3 x)$ and $0$ .

$f (\cos (3 x) + 2) = \frac{1}{3} \cdot \arccos (\cos (3 x))$

Step 5.3.4

Combine $\frac{1}{3}$ and $\arccos (\cos (3 x))$ .

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

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \cos (3 x) + 2$ is the inverse of $f (x) = \frac{1}{3} \cdot \arccos (x - 2)$ .

$f^{- 1} (x) = \cos (3 x) + 2$