Trigonometry Examples

$f (x) = 7 \arcsin (x^{2})$

Step 1

Write $f (x) = 7 \arcsin (x^{2})$ as an equation.

$y = 7 \arcsin (x^{2})$

Step 2

Interchange the variables.

$x = 7 \arcsin (y^{2})$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $7 \arcsin (y^{2}) = x$ .

$7 \arcsin (y^{2}) = x$

Step 3.2

Divide each term in $7 \arcsin (y^{2}) = x$ by $7$ and simplify.

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Step 3.2.1

Divide each term in $7 \arcsin (y^{2}) = x$ by $7$ .

$\frac{7 \arcsin (y^{2})}{7} = \frac{x}{7}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $7$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{7 \arcsin (y^{2})}{7} = \frac{x}{7}$

Step 3.2.2.1.2

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

$\arcsin (y^{2}) = \frac{x}{7}$

Step 3.3

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

$y^{2} = \sin (\frac{x}{7})$

Step 3.4

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

$y = \pm \sqrt{\sin (\frac{x}{7})}$

Step 3.5

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

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Step 3.5.1

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

$y = \sqrt{\sin (\frac{x}{7})}$

Step 3.5.2

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

$y = - \sqrt{\sin (\frac{x}{7})}$

Step 3.5.3

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

$y = \sqrt{\sin (\frac{x}{7})}$

$y = - \sqrt{\sin (\frac{x}{7})}$

$y = \sqrt{\sin (\frac{x}{7})}$

$y = - \sqrt{\sin (\frac{x}{7})}$

$y = \sqrt{\sin (\frac{x}{7})}$

$y = - \sqrt{\sin (\frac{x}{7})}$

Step 4

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

$f^{- 1} (x) = \sqrt{\sin (\frac{x}{7})}, - \sqrt{\sin (\frac{x}{7})}$

Step 5

Verify if $f^{- 1} (x) = \sqrt{\sin (\frac{x}{7})}, - \sqrt{\sin (\frac{x}{7})}$ is the inverse of $f (x) = 7 \arcsin (x^{2})$ .

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Step 5.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 7 \arcsin (x^{2})$ and $f^{- 1} (x) = \sqrt{\sin (\frac{x}{7})}, - \sqrt{\sin (\frac{x}{7})}$ and compare them.

Step 5.2

Find the range of $f (x) = 7 \arcsin (x^{2})$ .

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Step 5.2.1

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

Interval Notation:

$[- \frac{7 π}{2}, \frac{7 π}{2}]$

Step 5.3

Find the domain of the inverse.

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Step 5.3.1

Find the domain of $\sqrt{\sin (\frac{x}{7})}$ .

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Step 5.3.1.1

Set the radicand in $\sqrt{\sin (\frac{x}{7})}$ greater than or equal to $0$ to find where the expression is defined.

$\sin (\frac{x}{7}) \geq 0$

Step 5.3.1.2

Solve for $x$ .

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Step 5.3.1.2.1

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

$\frac{x}{7} \geq \arcsin (0)$

Step 5.3.1.2.2

Simplify the right side.

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Step 5.3.1.2.2.1

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

$\frac{x}{7} \geq 0$

Step 5.3.1.2.3

Multiply both sides by $7$ .

$\frac{x}{7} \cdot 7 \geq 0 \cdot 7$

Step 5.3.1.2.4

Simplify.

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Step 5.3.1.2.4.1

Simplify the left side.

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Step 5.3.1.2.4.1.1

Cancel the common factor of $7$ .

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Step 5.3.1.2.4.1.1.1

Cancel the common factor.

$\frac{x}{7} \cdot 7 \geq 0 \cdot 7$

Step 5.3.1.2.4.1.1.2

Rewrite the expression.

$x \geq 0 \cdot 7$

Step 5.3.1.2.4.2

Simplify the right side.

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Step 5.3.1.2.4.2.1

Multiply $0$ by $7$ .

$x \geq 0$

Step 5.3.1.2.5

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.

$\frac{x}{7} = π - 0$

Step 5.3.1.2.6

Solve for $x$ .

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Step 5.3.1.2.6.1

Multiply both sides of the equation by $7$ .

$7 \frac{x}{7} = 7 (π - 0)$

Step 5.3.1.2.6.2

Simplify both sides of the equation.

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Step 5.3.1.2.6.2.1

Simplify the left side.

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Step 5.3.1.2.6.2.1.1

Cancel the common factor of $7$ .

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Step 5.3.1.2.6.2.1.1.1

Cancel the common factor.

$7 \frac{x}{7} = 7 (π - 0)$

Step 5.3.1.2.6.2.1.1.2

Rewrite the expression.

$x = 7 (π - 0)$

Step 5.3.1.2.6.2.2

Simplify the right side.

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Step 5.3.1.2.6.2.2.1

Subtract $0$ from $π$ .

$x = 7 π$

Step 5.3.1.2.7

Find the period of $\sin (\frac{x}{7})$ .

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Step 5.3.1.2.7.1

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

$\frac{2 π}{| b |}$

Step 5.3.1.2.7.2

Replace $b$ with $\frac{1}{7}$ in the formula for period.

$\frac{2 π}{| \frac{1}{7} |}$

Step 5.3.1.2.7.3

$\frac{1}{7}$ is approximately $0.\overline{142857}$ which is positive so remove the absolute value

$\frac{2 π}{\frac{1}{7}}$

Step 5.3.1.2.7.4

Multiply the numerator by the reciprocal of the denominator.

$2 π \cdot 7$

Step 5.3.1.2.7.5

Multiply $7$ by $2$ .

$14 π$

Step 5.3.1.2.8

The period of the $\sin (\frac{x}{7})$ function is $14 π$ so values will repeat every $14 π$ radians in both directions.

$x = 14 π n, 7 π + 14 π n$ , for any integer $n$

Step 5.3.1.2.9

Consolidate the answers.

$x = 7 π n$ , for any integer $n$

Step 5.3.1.2.10

Use each root to create test intervals.

$0 < x < 7 π$

$7 π < x < 14 π$

Step 5.3.1.2.11

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 5.3.1.2.11.1

Test a value on the interval $0 < x < 7 π$ to see if it makes the inequality true.

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Step 5.3.1.2.11.1.1

Choose a value on the interval $0 < x < 7 π$ and see if this value makes the original inequality true.

$x = 2$

Step 5.3.1.2.11.1.2

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

$\sin (\frac{2}{7}) \geq 0$

Step 5.3.1.2.11.1.3

The left side $0.28184285$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 5.3.1.2.11.2

Test a value on the interval $7 π < x < 14 π$ to see if it makes the inequality true.

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Step 5.3.1.2.11.2.1

Choose a value on the interval $7 π < x < 14 π$ and see if this value makes the original inequality true.

$x = 24$

Step 5.3.1.2.11.2.2

Replace $x$ with $24$ in the original inequality.

$\sin (\frac{24}{7}) \geq 0$

Step 5.3.1.2.11.2.3

The left side $- 0.28305585$ is less than the right side $0$ , which means that the given statement is false.

False

Step 5.3.1.2.11.3

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

$0 < x < 7 π$ True

$7 π < x < 14 π$ False

$0 < x < 7 π$ True

$7 π < x < 14 π$ False

Step 5.3.1.2.12

The solution consists of all of the true intervals.

$0 + 14 π n \leq x \leq 7 π + 14 π n$ , for any integer $n$

Step 5.3.1.3

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

${x | x = 0 + 14 π n, x = 7 π + 14 π n}$ $n$ , for any integer $n$

Step 5.3.2

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Step 5.3.2.1

The union consists of all of the elements that are contained in each interval.

No solution

Step 5.4

Since the domain of $f^{- 1} (x) = \sqrt{\sin (\frac{x}{7})}, - \sqrt{\sin (\frac{x}{7})}$ is not equal to the range of $f (x) = 7 \arcsin (x^{2})$ , then $f^{- 1} (x) = \sqrt{\sin (\frac{x}{7})}, - \sqrt{\sin (\frac{x}{7})}$ is not an inverse of $f (x) = 7 \arcsin (x^{2})$ .

There is no inverse

Step 6