Trigonometry Examples

$f^{- 1} (81)$

Step 1

Interchange the variables.

$f = y^{- 1} (81)$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $y^{- 1} \cdot 81 = f$ .

$y^{- 1} \cdot 81 = f$

Step 2.2

Divide each term in $y^{- 1} \cdot 81 = f$ by $81$ and simplify.

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Step 2.2.1

Divide each term in $y^{- 1} \cdot 81 = f$ by $81$ .

$\frac{y^{- 1} \cdot 81}{81} = \frac{f}{81}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Move $y^{- 1}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{81}{81 y} = \frac{f}{81}$

Step 2.2.2.2

Cancel the common factor of $81$ .

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Step 2.2.2.2.1

Cancel the common factor.

$\frac{81}{81 y} = \frac{f}{81}$

Step 2.2.2.2.2

Rewrite the expression.

$\frac{1}{y} = \frac{f}{81}$

Step 2.3

Find the LCD of the terms in the equation.

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Step 2.3.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y, 81$

Step 2.3.2

Since $y, 81$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $1, 81$ then find LCM for the variable part $y^{1}$ .

Step 2.3.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 2.3.4

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

Step 2.3.5

The prime factors for $81$ are $3 \cdot 3 \cdot 3 \cdot 3$ .

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Step 2.3.5.1

$81$ has factors of $3$ and $27$ .

$3 \cdot 27$

Step 2.3.5.2

$27$ has factors of $3$ and $9$ .

$3 \cdot 3 \cdot 9$

Step 2.3.5.3

$9$ has factors of $3$ and $3$ .

$3 \cdot 3 \cdot 3 \cdot 3$

Step 2.3.6

Multiply $3 \cdot 3 \cdot 3 \cdot 3$ .

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Step 2.3.6.1

Multiply $3$ by $3$ .

$9 \cdot 3 \cdot 3$

Step 2.3.6.2

Multiply $9$ by $3$ .

$27 \cdot 3$

Step 2.3.6.3

Multiply $27$ by $3$ .

$81$

Step 2.3.7

The factor for $y^{1}$ is $y$ itself.

$y^{1} = y$

$y$ occurs $1$ time.

Step 2.3.8

The LCM of $y^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$y$

Step 2.3.9

The LCM for $y, 81$ is the numeric part $81$ multiplied by the variable part.

$81 y$

Step 2.4

Multiply each term in $\frac{1}{y} = \frac{f}{81}$ by $81 y$ to eliminate the fractions.

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Step 2.4.1

Multiply each term in $\frac{1}{y} = \frac{f}{81}$ by $81 y$ .

$\frac{1}{y} (81 y) = \frac{f}{81} (81 y)$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

Rewrite using the commutative property of multiplication.

$81 \frac{1}{y} y = \frac{f}{81} (81 y)$

Step 2.4.2.2

Combine $81$ and $\frac{1}{y}$ .

$\frac{81}{y} y = \frac{f}{81} (81 y)$

Step 2.4.2.3

Cancel the common factor of $y$ .

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Step 2.4.2.3.1

Cancel the common factor.

$\frac{81}{y} y = \frac{f}{81} (81 y)$

Step 2.4.2.3.2

Rewrite the expression.

$81 = \frac{f}{81} (81 y)$

Step 2.4.3

Simplify the right side.

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Step 2.4.3.1

Rewrite using the commutative property of multiplication.

$81 = 81 \frac{f}{81} y$

Step 2.4.3.2

Cancel the common factor of $81$ .

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Step 2.4.3.2.1

Cancel the common factor.

$81 = 81 \frac{f}{81} y$

Step 2.4.3.2.2

Rewrite the expression.

$81 = f y$

Step 2.5

Solve the equation.

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Step 2.5.1

Rewrite the equation as $f y = 81$ .

$f y = 81$

Step 2.5.2

Divide each term in $f y = 81$ by $f$ and simplify.

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Step 2.5.2.1

Divide each term in $f y = 81$ by $f$ .

$\frac{f y}{f} = \frac{81}{f}$

Step 2.5.2.2

Simplify the left side.

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Step 2.5.2.2.1

Cancel the common factor of $f$ .

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Step 2.5.2.2.1.1

Cancel the common factor.

$\frac{f y}{f} = \frac{81}{f}$

Step 2.5.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{81}{f}$

Step 3

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

$f^{- 1} (f) = \frac{81}{f}$

Step 4

Verify if $f^{- 1} (f) = \frac{81}{f}$ is the inverse of $f (f) = f^{- 1} (81)$ .

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

Evaluate $f^{- 1} (f^{- 1} (81))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (f^{- 1} (81)) = \frac{81}{f^{- 1} (81)}$

Step 4.2.3

Move $f^{- 1}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$f^{- 1} (f^{- 1} (81)) = \frac{81 f}{81}$

Step 4.2.4

Cancel the common factor of $81$ .

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Step 4.2.4.1

Cancel the common factor.

$f^{- 1} (f^{- 1} (81)) = \frac{81 f}{81}$

Step 4.2.4.2

Divide $f$ by $1$ .

$f^{- 1} (f^{- 1} (81)) = f$

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

Evaluate $f (\frac{81}{f})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{81}{f}) = {(\frac{81}{f})}^{- 1} (81)$

Step 4.3.3

Change the sign of the exponent by rewriting the base as its reciprocal.

$f (\frac{81}{f}) = \frac{f}{81} \cdot 81$

Step 4.3.4

Cancel the common factor of $81$ .

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Step 4.3.4.1

Cancel the common factor.

$f (\frac{81}{f}) = \frac{f}{81} \cdot 81$

Step 4.3.4.2

Rewrite the expression.

$f (\frac{81}{f}) = f$

Step 4.4

Since $f^{- 1} (f (f)) = f$ and $f (f^{- 1} (f)) = f$ , then $f^{- 1} (f) = \frac{81}{f}$ is the inverse of $f (f) = f^{- 1} (81)$ .

$f^{- 1} (f) = \frac{81}{f}$