Trigonometry Examples

$\cos (x - y)$

Step 1

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

$x - y = \arccos (0)$

Step 2

Simplify the right side.

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Step 2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x - y = \frac{π}{2}$

Step 3

Subtract $x$ from both sides of the equation.

$- y = \frac{π}{2} - x$

Step 4

Divide each term in $- y = \frac{π}{2} - x$ by $- 1$ and simplify.

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Step 4.1

Divide each term in $- y = \frac{π}{2} - x$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\frac{π}{2}}{- 1} + \frac{- x}{- 1}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\frac{π}{2}}{- 1} + \frac{- x}{- 1}$

Step 4.2.2

Divide $y$ by $1$ .

$y = \frac{\frac{π}{2}}{- 1} + \frac{- x}{- 1}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Move the negative one from the denominator of $\frac{\frac{π}{2}}{- 1}$ .

$y = - 1 \cdot \frac{π}{2} + \frac{- x}{- 1}$

Step 4.3.1.2

Rewrite $- 1 \cdot \frac{π}{2}$ as $- \frac{π}{2}$ .

$y = - \frac{π}{2} + \frac{- x}{- 1}$

Step 4.3.1.3

Dividing two negative values results in a positive value.

$y = - \frac{π}{2} + \frac{x}{1}$

Step 4.3.1.4

Divide $x$ by $1$ .

$y = - \frac{π}{2} + x$

Step 5

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x - y = 2 π - \frac{π}{2}$

Step 6

Solve for $y$ .

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Step 6.1

Simplify $2 π - \frac{π}{2}$ .

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Step 6.1.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 6.1.2

Combine fractions.

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Step 6.1.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 6.1.2.2

Combine the numerators over the common denominator.

$x - y = \frac{2 π \cdot 2 - π}{2}$

Step 6.1.3

Simplify the numerator.

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Step 6.1.3.1

Multiply $2$ by $2$ .

$x - y = \frac{4 π - π}{2}$

Step 6.1.3.2

Subtract $π$ from $4 π$ .

$x - y = \frac{3 π}{2}$

Step 6.2

Subtract $x$ from both sides of the equation.

$- y = \frac{3 π}{2} - x$

Step 6.3

Divide each term in $- y = \frac{3 π}{2} - x$ by $- 1$ and simplify.

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Step 6.3.1

Divide each term in $- y = \frac{3 π}{2} - x$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\frac{3 π}{2}}{- 1} + \frac{- x}{- 1}$

Step 6.3.2

Simplify the left side.

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Step 6.3.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\frac{3 π}{2}}{- 1} + \frac{- x}{- 1}$

Step 6.3.2.2

Divide $y$ by $1$ .

$y = \frac{\frac{3 π}{2}}{- 1} + \frac{- x}{- 1}$

Step 6.3.3

Simplify the right side.

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Step 6.3.3.1

Simplify each term.

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Step 6.3.3.1.1

Move the negative one from the denominator of $\frac{\frac{3 π}{2}}{- 1}$ .

$y = - 1 \cdot \frac{3 π}{2} + \frac{- x}{- 1}$

Step 6.3.3.1.2

Rewrite $- 1 \cdot \frac{3 π}{2}$ as $- \frac{3 π}{2}$ .

$y = - \frac{3 π}{2} + \frac{- x}{- 1}$

Step 6.3.3.1.3

Dividing two negative values results in a positive value.

$y = - \frac{3 π}{2} + \frac{x}{1}$

Step 6.3.3.1.4

Divide $x$ by $1$ .

$y = - \frac{3 π}{2} + x$

Step 7

Interchange the variables.

$x = - \frac{3 π}{2} + y$

Step 8

Solve for $y$ .

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Step 8.1

Rewrite the equation as $- \frac{3 π}{2} + y = x$ .

$- \frac{3 π}{2} + y = x$

Step 8.2

Add $\frac{3 π}{2}$ to both sides of the equation.

$y = x + \frac{3 π}{2}$

Step 9

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

$f^{- 1} (x) = x + \frac{3 π}{2}$

Step 10

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

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Step 10.1

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

Step 10.2

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

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Step 10.2.1

Set up the composite result function.

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

Step 10.2.2

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

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

Step 10.2.3

Combine the opposite terms in $(- \frac{3 π}{2} + x) + \frac{3 π}{2}$ .

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Step 10.2.3.1

Add $- \frac{3 π}{2}$ and $\frac{3 π}{2}$ .

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

Step 10.2.3.2

Add $x$ and $0$ .

$f^{- 1} (- \frac{3 π}{2} + x) = x$

Step 10.3

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

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Step 10.3.1

Set up the composite result function.

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

Step 10.3.2

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

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

Step 10.3.3

Remove parentheses.

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

Step 10.3.4

Combine the opposite terms in $- \frac{3 π}{2} + x + \frac{3 π}{2}$ .

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Step 10.3.4.1

Add $- \frac{3 π}{2}$ and $\frac{3 π}{2}$ .

$f (x + \frac{3 π}{2}) = x + 0$

Step 10.3.4.2

Add $x$ and $0$ .

$f (x + \frac{3 π}{2}) = x$

Step 10.4

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

$f^{- 1} (x) = x + \frac{3 π}{2}$