Trigonometry Examples

$y = \frac{1}{2} \arccos (π x) - 3$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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Step 2.1

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

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

Step 2.2

Simplify the left side.

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Step 2.2.1

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

$\frac{\arccos (π y)}{2} - 3 = x$

Step 2.3

Add $3$ to both sides of the equation.

$\frac{\arccos (π y)}{2} = x + 3$

Step 2.4

Multiply both sides by $2$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Simplify the left side.

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Step 2.5.1.1

Cancel the common factor of $2$ .

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Step 2.5.1.1.1

Cancel the common factor.

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

Step 2.5.1.1.2

Rewrite the expression.

$\arccos (π y) = (x + 3) \cdot 2$

Step 2.5.2

Simplify the right side.

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Step 2.5.2.1

Simplify $(x + 3) \cdot 2$ .

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Step 2.5.2.1.1

Apply the distributive property.

$\arccos (π y) = x \cdot 2 + 3 \cdot 2$

Step 2.5.2.1.2

Simplify the expression.

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Step 2.5.2.1.2.1

Move $2$ to the left of $x$ .

$\arccos (π y) = 2 \cdot x + 3 \cdot 2$

Step 2.5.2.1.2.2

Multiply $3$ by $2$ .

$\arccos (π y) = 2 x + 6$

Step 2.6

Solve for $y$ .

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Step 2.6.1

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

$π y = \cos (2 x + 6)$

Step 2.6.2

Divide each term in $π y = \cos (2 x + 6)$ by $π$ and simplify.

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Step 2.6.2.1

Divide each term in $π y = \cos (2 x + 6)$ by $π$ .

$\frac{π y}{π} = \frac{\cos (2 x + 6)}{π}$

Step 2.6.2.2

Simplify the left side.

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Step 2.6.2.2.1

Cancel the common factor of $π$ .

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Step 2.6.2.2.1.1

Cancel the common factor.

$\frac{π y}{π} = \frac{\cos (2 x + 6)}{π}$

Step 2.6.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{\cos (2 x + 6)}{π}$

Step 3

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

$f^{- 1} (x) = \frac{\cos (2 x + 6)}{π}$

Step 4

Verify if $f^{- 1} (x) = \frac{\cos (2 x + 6)}{π}$ is the inverse of $f (x) = \frac{1}{2} \cdot \arccos (π x) - 3$ .

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

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

$f^{- 1} (\frac{1}{2} \cdot \arccos (π x) - 3) = \frac{\cos (2 (\frac{1}{2} \cdot \arccos (π x) - 3) + 6)}{π}$

Step 4.2.3

Simplify the numerator.

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Step 4.2.3.1

Simplify each term.

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Step 4.2.3.1.1

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

$f^{- 1} (\frac{1}{2} \cdot \arccos (π x) - 3) = \frac{\cos (2 (\frac{\arccos (π x)}{2} - 3) + 6)}{π}$

Step 4.2.3.1.2

Apply the distributive property.

$f^{- 1} (\frac{1}{2} \cdot \arccos (π x) - 3) = \frac{\cos (2 (\frac{\arccos (π x)}{2}) + 2 \cdot - 3 + 6)}{π}$

Step 4.2.3.1.3

Cancel the common factor of $2$ .

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Step 4.2.3.1.3.1

Cancel the common factor.

$f^{- 1} (\frac{1}{2} \cdot \arccos (π x) - 3) = \frac{\cos (2 (\frac{\arccos (π x)}{2}) + 2 \cdot - 3 + 6)}{π}$

Step 4.2.3.1.3.2

Rewrite the expression.

$f^{- 1} (\frac{1}{2} \cdot \arccos (π x) - 3) = \frac{\cos (\arccos (π x) + 2 \cdot - 3 + 6)}{π}$

Step 4.2.3.1.4

Multiply $2$ by $- 3$ .

$f^{- 1} (\frac{1}{2} \cdot \arccos (π x) - 3) = \frac{\cos (\arccos (π x) - 6 + 6)}{π}$

Step 4.2.3.2

Combine the opposite terms in $\arccos (π x) - 6 + 6$ .

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Step 4.2.3.2.1

Add $- 6$ and $6$ .

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

Step 4.2.3.2.2

Add $\arccos (π x)$ and $0$ .

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

Step 4.2.3.3

The functions cosine and arccosine are inverses.

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

Step 4.2.4

Cancel the common factor of $π$ .

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Step 4.2.4.1

Cancel the common factor.

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

Step 4.2.4.2

Divide $x$ by $1$ .

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

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

Evaluate $f (\frac{\cos (2 x + 6)}{π})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\cos (2 x + 6)}{π}) = \frac{1}{2} \cdot \arccos (π (\frac{\cos (2 x + 6)}{π})) - 3$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Cancel the common factor of $π$ .

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Step 4.3.3.1.1

Cancel the common factor.

$f (\frac{\cos (2 x + 6)}{π}) = \frac{1}{2} \cdot \arccos (π (\frac{\cos (2 x + 6)}{π})) - 3$

Step 4.3.3.1.2

Rewrite the expression.

$f (\frac{\cos (2 x + 6)}{π}) = \frac{1}{2} \cdot \arccos (\cos (2 x + 6)) - 3$

Step 4.3.3.2

Combine $\frac{1}{2}$ and $\arccos (\cos (2 x + 6))$ .

$f (\frac{\cos (2 x + 6)}{π}) = \frac{\arccos (\cos (2 x + 6))}{2} - 3$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\cos (2 x + 6)}{π}$ is the inverse of $f (x) = \frac{1}{2} \cdot \arccos (π x) - 3$ .

$f^{- 1} (x) = \frac{\cos (2 x + 6)}{π}$