Trigonometry Examples

$\frac{\sin (x + y)}{\sin (x - y)}$

Step 1

Set the numerator equal to zero.

$\sin (x + y) = 0$

Step 2

Solve the equation for $y$ .

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Step 2.1

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

$x + y = \arcsin (0)$

Step 2.2

Simplify the right side.

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Step 2.2.1

The exact value of $\arcsin (0)$ is $0$ .

$x + y = 0$

Step 2.3

Subtract $x$ from both sides of the equation.

$y = - x$

Step 2.4

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

$x + y = π - 0$

Step 2.5

Solve for $y$ .

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Step 2.5.1

Subtract $0$ from $π$ .

$x + y = π$

Step 2.5.2

Subtract $x$ from both sides of the equation.

$y = π - x$

Step 2.6

Subtract $x$ from both sides of the equation.

$y = - x$

Step 3

Interchange the variables.

$x = - y$

Step 4

Solve for $y$ .

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Step 4.1

Rewrite the equation as $- y = x$ .

$- y = x$

Step 4.2

Divide each term in $- y = x$ by $- 1$ and simplify.

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Step 4.2.1

Divide each term in $- y = x$ by $- 1$ .

$\frac{- y}{- 1} = \frac{x}{- 1}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{x}{- 1}$

Step 4.2.2.2

Divide $y$ by $1$ .

$y = \frac{x}{- 1}$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

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

$y = - 1 \cdot x$

Step 4.2.3.2

Rewrite $- 1 \cdot x$ as $- x$ .

$y = - x$

Step 5

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

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

Step 6

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

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Step 6.1

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

Step 6.2

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

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Step 6.2.1

Set up the composite result function.

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

Step 6.2.2

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

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

Step 6.2.3

Multiply $- (- x)$ .

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Step 6.2.3.1

Multiply $- 1$ by $- 1$ .

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

Step 6.2.3.2

Multiply $x$ by $1$ .

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

Step 6.3

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

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Step 6.3.1

Set up the composite result function.

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

Step 6.3.2

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

$f (- x) = - (- x)$

Step 6.3.3

Multiply $- (- x)$ .

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Step 6.3.3.1

Multiply $- 1$ by $- 1$ .

$f (- x) = 1 x$

Step 6.3.3.2

Multiply $x$ by $1$ .

$f (- x) = x$

Step 6.4

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

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