Algebra Examples

$f (x) = \arcsin (\sin (x))$

Step 1

Set the argument in $\arcsin (\sin (x))$ greater than or equal to $- 1$ to find where the expression is defined.

$\sin (x) \geq - 1$

Step 2

Solve for $x$ .

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Step 2.1

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

$x \geq \arcsin (- 1)$

Step 2.2

Simplify the right side.

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Step 2.2.1

The exact value of $\arcsin (- 1)$ is $- \frac{π}{2}$ .

$x \geq - \frac{π}{2}$

Step 2.3

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 2.4

Simplify the expression to find the second solution.

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Step 2.4.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 2.4.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 2.5

Find the period of $\sin (x)$ .

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Step 2.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 2.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 2.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.6

Add $2 π$ to every negative angle to get positive angles.

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Step 2.6.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 2.6.2

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 2.6.3

Combine fractions.

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Step 2.6.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 2.6.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 2.6.4

Simplify the numerator.

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Step 2.6.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 2.6.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 2.6.5

List the new angles.

$x = \frac{3 π}{2}$

Step 2.7

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{3 π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 2.8

Consolidate the answers.

$x = \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 2.9

Use each root to create test intervals.

$\frac{3 π}{2} < x < \frac{7 π}{2}$

Step 2.10

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 2.10.1

Test a value on the interval $\frac{3 π}{2} < x < \frac{7 π}{2}$ to see if it makes the inequality true.

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Step 2.10.1.1

Choose a value on the interval $\frac{3 π}{2} < x < \frac{7 π}{2}$ and see if this value makes the original inequality true.

$x = 8$

Step 2.10.1.2

Replace $x$ with $8$ in the original inequality.

$\sin (8) \geq - 1$

Step 2.10.1.3

The left side $0.98935824$ is greater than the right side $- 1$ , which means that the given statement is always true.

True

Step 2.10.2

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

$\frac{3 π}{2} < x < \frac{7 π}{2}$ True

Step 2.11

The solution consists of all of the true intervals.

$\frac{3 π}{2} + 2 π n \leq x \leq \frac{7 π}{2} + 2 π n$ , for any integer $n$

Step 3

Set the argument in $\arcsin (\sin (x))$ less than or equal to $1$ to find where the expression is defined.

$\sin (x) \leq 1$

Step 4

Solve for $x$ .

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Step 4.1

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

$x \leq \arcsin (1)$

Step 4.2

Simplify the right side.

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Step 4.2.1

The exact value of $\arcsin (1)$ is $\frac{π}{2}$ .

$x \leq \frac{π}{2}$

Step 4.3

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.

$x = π - \frac{π}{2}$

Step 4.4

Simplify $π - \frac{π}{2}$ .

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Step 4.4.1

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

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

Step 4.4.2

Combine fractions.

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Step 4.4.2.1

Combine $π$ and $\frac{2}{2}$ .

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

Step 4.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 4.4.3

Simplify the numerator.

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Step 4.4.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 4.4.3.2

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Step 4.5

Find the period of $\sin (x)$ .

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Step 4.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 4.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.6

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{2} + 2 π n$ , for any integer $n$

Step 4.7

Use each root to create test intervals.

$\frac{π}{2} < x < \frac{5 π}{2}$

Step 4.8

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 4.8.1

Test a value on the interval $\frac{π}{2} < x < \frac{5 π}{2}$ to see if it makes the inequality true.

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Step 4.8.1.1

Choose a value on the interval $\frac{π}{2} < x < \frac{5 π}{2}$ and see if this value makes the original inequality true.

$x = 5$

Step 4.8.1.2

Replace $x$ with $5$ in the original inequality.

$\sin (5) \leq 1$

Step 4.8.1.3

The left side $- 0.95892427$ is less than the right side $1$ , which means that the given statement is always true.

True

Step 4.8.2

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

$\frac{π}{2} < x < \frac{5 π}{2}$ True

Step 4.9

The solution consists of all of the true intervals.

$\frac{π}{2} + 2 π n \leq x \leq \frac{5 π}{2} + 2 π n$ , for any integer $n$

Step 5

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

Set-Builder Notation:

${x | x = \frac{3 π}{2} + 2 π n, x = \frac{7 π}{2} + 2 π n, x = \frac{π}{2} + 2 π n, x = \frac{5 π}{2} + 2 π n}$ , for any integer $n$

Step 6

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

Interval Notation:

$[- \frac{π}{2}, \frac{π}{2}]$

Set-Builder Notation:

${y | - \frac{π}{2} \leq y \leq \frac{π}{2}}$

Step 7

Determine the domain and range.

Domain: ${x | x = \frac{3 π}{2} + 2 π n, x = \frac{7 π}{2} + 2 π n, x = \frac{π}{2} + 2 π n, x = \frac{5 π}{2} + 2 π n}$ , for any integer $n$

Range: $[- \frac{π}{2}, \frac{π}{2}], {y | - \frac{π}{2} \leq y \leq \frac{π}{2}}$

Step 8