Trigonometry Examples

$1 - \sec^{2} (x)$

Step 1

Interchange the variables.

$x = 1 - \sec^{2} (y)$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $1 - \sec^{2} (y) = x$ .

$1 - \sec^{2} (y) = x$

Step 2.2

Subtract $1$ from both sides of the equation.

$- \sec^{2} (y) = x - 1$

Step 2.3

Divide each term in $- \sec^{2} (y) = x - 1$ by $- 1$ and simplify.

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Step 2.3.1

Divide each term in $- \sec^{2} (y) = x - 1$ by $- 1$ .

$\frac{- \sec^{2} (y)}{- 1} = \frac{x}{- 1} + \frac{- 1}{- 1}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Dividing two negative values results in a positive value.

$\frac{\sec^{2} (y)}{1} = \frac{x}{- 1} + \frac{- 1}{- 1}$

Step 2.3.2.2

Divide $\sec^{2} (y)$ by $1$ .

$\sec^{2} (y) = \frac{x}{- 1} + \frac{- 1}{- 1}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Simplify each term.

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Step 2.3.3.1.1

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

$\sec^{2} (y) = - 1 \cdot x + \frac{- 1}{- 1}$

Step 2.3.3.1.2

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

$\sec^{2} (y) = - x + \frac{- 1}{- 1}$

Step 2.3.3.1.3

Divide $- 1$ by $- 1$ .

$\sec^{2} (y) = - x + 1$

Step 2.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\sec (y) = \pm \sqrt{- x + 1}$

Step 2.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.5.1

First, use the positive value of the $\pm$ to find the first solution.

$\sec (y) = \sqrt{- x + 1}$

Step 2.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$\sec (y) = - \sqrt{- x + 1}$

Step 2.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$\sec (y) = \sqrt{- x + 1}, - \sqrt{- x + 1}$

Step 2.6

Set up each of the solutions to solve for $y$ .

$\sec (y) = \sqrt{- x + 1}$

$\sec (y) = - \sqrt{- x + 1}$

Step 2.7

Solve for $y$ in $\sec (y) = \sqrt{- x + 1}$ .

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Step 2.7.1

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

$y = arcsec (\sqrt{- x + 1})$

Step 2.8

Solve for $y$ in $\sec (y) = - \sqrt{- x + 1}$ .

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Step 2.8.1

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

$y = arcsec (- \sqrt{- x + 1})$

Step 2.9

List all of the solutions.

$y = arcsec (\sqrt{- x + 1})$

$y = arcsec (- \sqrt{- x + 1})$

$y = arcsec (\sqrt{- x + 1})$

$y = arcsec (- \sqrt{- x + 1})$

Step 3

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

$f^{- 1} (x) = arcsec (\sqrt{- x + 1}), arcsec (- \sqrt{- x + 1})$

Step 4

Verify if $f^{- 1} (x) = arcsec (\sqrt{- x + 1}), arcsec (- \sqrt{- x + 1})$ is the inverse of $f (x) = 1 - \sec^{2} (x)$ .

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Step 4.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 1 - \sec^{2} (x)$ and $f^{- 1} (x) = arcsec (\sqrt{- x + 1}), arcsec (- \sqrt{- x + 1})$ and compare them.

Step 4.2

Find the range of $f (x) = 1 - \sec^{2} (x)$ .

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Step 4.2.1

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[2, \infty)$

Step 4.3

Find the domain of $arcsec (\sqrt{- x + 1})$ .

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Step 4.3.1

Set the radicand in $\sqrt{- x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$- x + 1 \geq 0$

Step 4.3.2

Solve for $x$ .

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Step 4.3.2.1

Subtract $1$ from both sides of the inequality.

$- x \geq - 1$

Step 4.3.2.2

Divide each term in $- x \geq - 1$ by $- 1$ and simplify.

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Step 4.3.2.2.1

Divide each term in $- x \geq - 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 1}{- 1}$

Step 4.3.2.2.2

Simplify the left side.

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Step 4.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 1}{- 1}$

Step 4.3.2.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{- 1}{- 1}$

Step 4.3.2.2.3

Simplify the right side.

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Step 4.3.2.2.3.1

Divide $- 1$ by $- 1$ .

$x \leq 1$

Step 4.3.3

Set the argument in $arcsec (\sqrt{- x + 1})$ less than or equal to $- 1$ to find where the expression is defined.

$\sqrt{- x + 1} \leq - 1$

Step 4.3.4

Solve for $x$ .

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Step 4.3.4.1

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{- x + 1}}^{2} \leq {(- 1)}^{2}$

Step 4.3.4.2

Simplify each side of the inequality.

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Step 4.3.4.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{- x + 1}$ as ${(- x + 1)}^{\frac{1}{2}}$ .

${({(- x + 1)}^{\frac{1}{2}})}^{2} \leq {(- 1)}^{2}$

Step 4.3.4.2.2

Simplify the left side.

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Step 4.3.4.2.2.1

Simplify ${({(- x + 1)}^{\frac{1}{2}})}^{2}$ .

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Step 4.3.4.2.2.1.1

Multiply the exponents in ${({(- x + 1)}^{\frac{1}{2}})}^{2}$ .

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Step 4.3.4.2.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(- x + 1)}^{\frac{1}{2} \cdot 2} \leq {(- 1)}^{2}$

Step 4.3.4.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 4.3.4.2.2.1.1.2.1

Cancel the common factor.

${(- x + 1)}^{\frac{1}{2} \cdot 2} \leq {(- 1)}^{2}$

Step 4.3.4.2.2.1.1.2.2

Rewrite the expression.

${(- x + 1)}^{1} \leq {(- 1)}^{2}$

Step 4.3.4.2.2.1.2

Simplify.

$- x + 1 \leq {(- 1)}^{2}$

Step 4.3.4.2.3

Simplify the right side.

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Step 4.3.4.2.3.1

Raise $- 1$ to the power of $2$ .

$- x + 1 \leq 1$

Step 4.3.4.3

Solve for $x$ .

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Step 4.3.4.3.1

Move all terms not containing $x$ to the right side of the inequality.

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Step 4.3.4.3.1.1

Subtract $1$ from both sides of the inequality.

$- x \leq 1 - 1$

Step 4.3.4.3.1.2

Subtract $1$ from $1$ .

$- x \leq 0$

Step 4.3.4.3.2

Divide each term in $- x \leq 0$ by $- 1$ and simplify.

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Step 4.3.4.3.2.1

Divide each term in $- x \leq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \geq \frac{0}{- 1}$

Step 4.3.4.3.2.2

Simplify the left side.

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Step 4.3.4.3.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \geq \frac{0}{- 1}$

Step 4.3.4.3.2.2.2

Divide $x$ by $1$ .

$x \geq \frac{0}{- 1}$

Step 4.3.4.3.2.3

Simplify the right side.

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Step 4.3.4.3.2.3.1

Divide $0$ by $- 1$ .

$x \geq 0$

Step 4.3.4.4

Find the domain of $\sqrt{- x + 1} + 1$ .

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Step 4.3.4.4.1

Set the radicand in $\sqrt{- x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$- x + 1 \geq 0$

Step 4.3.4.4.2

Solve for $x$ .

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Step 4.3.4.4.2.1

Subtract $1$ from both sides of the inequality.

$- x \geq - 1$

Step 4.3.4.4.2.2

Divide each term in $- x \geq - 1$ by $- 1$ and simplify.

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Step 4.3.4.4.2.2.1

Divide each term in $- x \geq - 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 1}{- 1}$

Step 4.3.4.4.2.2.2

Simplify the left side.

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Step 4.3.4.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 1}{- 1}$

Step 4.3.4.4.2.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{- 1}{- 1}$

Step 4.3.4.4.2.2.3

Simplify the right side.

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Step 4.3.4.4.2.2.3.1

Divide $- 1$ by $- 1$ .

$x \leq 1$

Step 4.3.4.4.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, 1]$

Step 4.3.4.5

Use each root to create test intervals.

$x < 0$

$0 < x < 1$

$x > 1$

Step 4.3.4.6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 4.3.4.6.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 4.3.4.6.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 4.3.4.6.1.2

Replace $x$ with $- 2$ in the original inequality.

$\sqrt{- (- 2) + 1} \leq - 1$

Step 4.3.4.6.1.3

The left side $1.7320508$ is greater than the right side $- 1$ , which means that the given statement is false.

False

Step 4.3.4.6.2

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

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Step 4.3.4.6.2.1

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Step 4.3.4.6.2.2

Replace $x$ with $0.5$ in the original inequality.

$\sqrt{- (0.5) + 1} \leq - 1$

Step 4.3.4.6.2.3

The left side $0.70710678$ is greater than the right side $- 1$ , which means that the given statement is false.

False

Step 4.3.4.6.3

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 4.3.4.6.3.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 4.3.4.6.3.2

Replace $x$ with $4$ in the original inequality.

$\sqrt{- (4) + 1} \leq - 1$

Step 4.3.4.6.3.3

The left side is not equal to the right side, which means that the given statement is false.

False

Step 4.3.4.6.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < 0$ False

$0 < x < 1$ False

$x > 1$ False

$x < 0$ False

$0 < x < 1$ False

$x > 1$ False

Step 4.3.4.7

Since there are no numbers that fall within the interval, this inequality has no solution.

No solution

Step 4.3.5

Set the argument in $arcsec (\sqrt{- x + 1})$ greater than or equal to $1$ to find where the expression is defined.

$\sqrt{- x + 1} \geq 1$

Step 4.3.6

Solve for $x$ .

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Step 4.3.6.1

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{- x + 1}}^{2} \geq 1^{2}$

Step 4.3.6.2

Simplify each side of the inequality.

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Step 4.3.6.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{- x + 1}$ as ${(- x + 1)}^{\frac{1}{2}}$ .

${({(- x + 1)}^{\frac{1}{2}})}^{2} \geq 1^{2}$

Step 4.3.6.2.2

Simplify the left side.

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Step 4.3.6.2.2.1

Simplify ${({(- x + 1)}^{\frac{1}{2}})}^{2}$ .

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Step 4.3.6.2.2.1.1

Multiply the exponents in ${({(- x + 1)}^{\frac{1}{2}})}^{2}$ .

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Step 4.3.6.2.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(- x + 1)}^{\frac{1}{2} \cdot 2} \geq 1^{2}$

Step 4.3.6.2.2.1.1.2

Cancel the common factor of $2$ .

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Step 4.3.6.2.2.1.1.2.1

Cancel the common factor.

${(- x + 1)}^{\frac{1}{2} \cdot 2} \geq 1^{2}$

Step 4.3.6.2.2.1.1.2.2

Rewrite the expression.

${(- x + 1)}^{1} \geq 1^{2}$

Step 4.3.6.2.2.1.2

Simplify.

$- x + 1 \geq 1^{2}$

Step 4.3.6.2.3

Simplify the right side.

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Step 4.3.6.2.3.1

One to any power is one.

$- x + 1 \geq 1$

Step 4.3.6.3

Solve for $x$ .

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Step 4.3.6.3.1

Move all terms not containing $x$ to the right side of the inequality.

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Step 4.3.6.3.1.1

Subtract $1$ from both sides of the inequality.

$- x \geq 1 - 1$

Step 4.3.6.3.1.2

Subtract $1$ from $1$ .

$- x \geq 0$

Step 4.3.6.3.2

Divide each term in $- x \geq 0$ by $- 1$ and simplify.

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Step 4.3.6.3.2.1

Divide each term in $- x \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{0}{- 1}$

Step 4.3.6.3.2.2

Simplify the left side.

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Step 4.3.6.3.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{0}{- 1}$

Step 4.3.6.3.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{0}{- 1}$

Step 4.3.6.3.2.3

Simplify the right side.

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Step 4.3.6.3.2.3.1

Divide $0$ by $- 1$ .

$x \leq 0$

Step 4.3.6.4

Find the domain of $\sqrt{- x + 1} - 1$ .

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Step 4.3.6.4.1

Set the radicand in $\sqrt{- x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$- x + 1 \geq 0$

Step 4.3.6.4.2

Solve for $x$ .

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Step 4.3.6.4.2.1

Subtract $1$ from both sides of the inequality.

$- x \geq - 1$

Step 4.3.6.4.2.2

Divide each term in $- x \geq - 1$ by $- 1$ and simplify.

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Step 4.3.6.4.2.2.1

Divide each term in $- x \geq - 1$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 1}{- 1}$

Step 4.3.6.4.2.2.2

Simplify the left side.

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Step 4.3.6.4.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 1}{- 1}$

Step 4.3.6.4.2.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{- 1}{- 1}$

Step 4.3.6.4.2.2.3

Simplify the right side.

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Step 4.3.6.4.2.2.3.1

Divide $- 1$ by $- 1$ .

$x \leq 1$

Step 4.3.6.4.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, 1]$

Step 4.3.6.5

The solution consists of all of the true intervals.

$x \leq 0$

Step 4.3.7

The domain is all values of $x$ that make the expression defined.

$(- \infty, 1]$

Step 4.4

Since the domain of $f^{- 1} (x) = arcsec (\sqrt{- x + 1}), arcsec (- \sqrt{- x + 1})$ is not equal to the range of $f (x) = 1 - \sec^{2} (x)$ , then $f^{- 1} (x) = arcsec (\sqrt{- x + 1}), arcsec (- \sqrt{- x + 1})$ is not an inverse of $f (x) = 1 - \sec^{2} (x)$ .

There is no inverse

Step 5