Trigonometry Examples

$f (x) = 2 + \cos (\frac{π}{2} x)$

Step 1

Write $f (x) = 2 + \cos (\frac{π}{2} x)$ as an equation.

$y = 2 + \cos (\frac{π}{2} x)$

Step 2

Interchange the variables.

$x = 2 + \cos (\frac{π}{2} y)$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $2 + \cos (\frac{π}{2} y) = x$ .

$2 + \cos (\frac{π}{2} y) = x$

Step 3.2

Subtract $2$ from both sides of the equation.

$\cos (\frac{π}{2} y) = x - 2$

Step 3.3

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

$\frac{π}{2} y = \arccos (x - 2)$

Step 3.4

Simplify the left side.

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Step 3.4.1

Combine $\frac{π}{2}$ and $y$ .

$\frac{π y}{2} = \arccos (x - 2)$

Step 3.5

Multiply both sides of the equation by $\frac{2}{π}$ .

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

Step 3.6

Simplify both sides of the equation.

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Step 3.6.1

Simplify the left side.

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Step 3.6.1.1

Simplify $\frac{2}{π} \cdot \frac{π y}{2}$ .

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Step 3.6.1.1.1

Cancel the common factor of $2$ .

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Step 3.6.1.1.1.1

Cancel the common factor.

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

Step 3.6.1.1.1.2

Rewrite the expression.

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

Step 3.6.1.1.2

Cancel the common factor of $π$ .

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Step 3.6.1.1.2.1

Factor $π$ out of $π y$ .

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

Step 3.6.1.1.2.2

Cancel the common factor.

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

Step 3.6.1.1.2.3

Rewrite the expression.

$y = \frac{2}{π} \arccos (x - 2)$

Step 3.6.2

Simplify the right side.

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Step 3.6.2.1

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

$y = \frac{2 \arccos (x - 2)}{π}$

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Simplify the numerator.

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Step 5.2.3.1

Combine the opposite terms in $2 + \cos (\frac{π}{2} x) - 2$ .

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Step 5.2.3.1.1

Subtract $2$ from $2$ .

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

Step 5.2.3.1.2

Add $0$ and $\cos (\frac{π}{2} x)$ .

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

Step 5.2.3.2

Combine $\frac{π}{2}$ and $x$ .

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

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

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Cancel the common factor of $π$ .

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Step 5.3.3.1.1

Cancel the common factor.

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

Step 5.3.3.1.2

Rewrite the expression.

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

Step 5.3.3.2

Cancel the common factor of $2$ .

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Step 5.3.3.2.1

Factor $2$ out of $2 \arccos (x - 2)$ .

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

Step 5.3.3.2.2

Cancel the common factor.

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

Step 5.3.3.2.3

Rewrite the expression.

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

Step 5.3.3.3

The functions cosine and arccosine are inverses.

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

Step 5.3.4

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

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Step 5.3.4.1

Subtract $2$ from $2$ .

$f (\frac{2 \arccos (x - 2)}{π}) = x + 0$

Step 5.3.4.2

Add $x$ and $0$ .

$f (\frac{2 \arccos (x - 2)}{π}) = x$

Step 5.4

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

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