Trigonometry Examples

$f (x) = \cos (x - 9) - 5$

Step 1

Write $f (x) = \cos (x - 9) - 5$ as an equation.

$y = \cos (x - 9) - 5$

Step 2

Interchange the variables.

$x = \cos (y - 9) - 5$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\cos (y - 9) - 5 = x$ .

$\cos (y - 9) - 5 = x$

Step 3.2

Add $5$ to both sides of the equation.

$\cos (y - 9) = x + 5$

Step 3.3

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

$y - 9 = \arccos (x + 5)$

Step 3.4

Add $9$ to both sides of the equation.

$y = \arccos (x + 5) + 9$

Step 4

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

$f^{- 1} (x) = \arccos (x + 5) + 9$

Step 5

Verify if $f^{- 1} (x) = \arccos (x + 5) + 9$ is the inverse of $f (x) = \cos (x - 9) - 5$ .

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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 $f^{- 1} (\cos (x - 9) - 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\cos (x - 9) - 5) = \arccos ((\cos (x - 9) - 5) + 5) + 9$

Step 5.2.3

Combine the opposite terms in $\arccos ((\cos (x - 9) - 5) + 5) + 9$ .

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Step 5.2.3.1

Add $- 5$ and $5$ .

$f^{- 1} (\cos (x - 9) - 5) = \arccos (\cos (x - 9) + 0) + 9$

Step 5.2.3.2

Add $\cos (x - 9)$ and $0$ .

$f^{- 1} (\cos (x - 9) - 5) = \arccos (\cos (x - 9)) + 9$

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 $f (\arccos (x + 5) + 9)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\arccos (x + 5) + 9) = \cos ((\arccos (x + 5) + 9) - 9) - 5$

Step 5.3.3

Combine the opposite terms in $\cos ((\arccos (x + 5) + 9) - 9) - 5$ .

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Step 5.3.3.1

Subtract $9$ from $9$ .

$f (\arccos (x + 5) + 9) = \cos (\arccos (x + 5) + 0) - 5$

Step 5.3.3.2

Add $\arccos (x + 5)$ and $0$ .

$f (\arccos (x + 5) + 9) = \cos (\arccos (x + 5)) - 5$

Step 5.3.4

The functions cosine and arccosine are inverses.

$f (\arccos (x + 5) + 9) = x + 5 - 5$

Step 5.3.5

Combine the opposite terms in $x + 5 - 5$ .

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Step 5.3.5.1

Subtract $5$ from $5$ .

$f (\arccos (x + 5) + 9) = x + 0$

Step 5.3.5.2

Add $x$ and $0$ .

$f (\arccos (x + 5) + 9) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \arccos (x + 5) + 9$ is the inverse of $f (x) = \cos (x - 9) - 5$ .

$f^{- 1} (x) = \arccos (x + 5) + 9$