Trigonometry Examples

$y = \sin (\sqrt{x}) + 2$

Step 1

Interchange the variables.

$x = \sin (\sqrt{y}) + 2$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\sin (\sqrt{y}) + 2 = x$ .

$\sin (\sqrt{y}) + 2 = x$

Step 2.2

Subtract $2$ from both sides of the equation.

$\sin (\sqrt{y}) = x - 2$

Step 2.3

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

$\sqrt{y} = \arcsin (x - 2)$

Step 2.4

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

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

Step 2.5

Simplify each side of the equation.

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Step 2.5.1

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

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

Step 2.5.2

Simplify the left side.

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Step 2.5.2.1

Simplify ${(y^{\frac{1}{2}})}^{2}$ .

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Step 2.5.2.1.1

Multiply the exponents in ${(y^{\frac{1}{2}})}^{2}$ .

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Step 2.5.2.1.1.1

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

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

Step 2.5.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.5.2.1.1.2.1

Cancel the common factor.

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

Step 2.5.2.1.1.2.2

Rewrite the expression.

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

Step 2.5.2.1.2

Simplify.

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

Step 3

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

$f^{- 1} (x) = {\arcsin (x - 2)}^{2}$

Step 4

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

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

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

$f^{- 1} (\sin (\sqrt{x}) + 2) = {\arcsin ((\sin (\sqrt{x}) + 2) - 2)}^{2}$

Step 4.2.3

Combine the opposite terms in $(\sin (\sqrt{x}) + 2) - 2$ .

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Step 4.2.3.1

Subtract $2$ from $2$ .

$f^{- 1} (\sin (\sqrt{x}) + 2) = {\arcsin (\sin (\sqrt{x}) + 0)}^{2}$

Step 4.2.3.2

Add $\sin (\sqrt{x})$ and $0$ .

$f^{- 1} (\sin (\sqrt{x}) + 2) = {\arcsin (\sin (\sqrt{x}))}^{2}$

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

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

$f ({\arcsin (x - 2)}^{2}) = \sin (\sqrt{{\arcsin (x - 2)}^{2}}) + 2$

Step 4.3.3

Remove parentheses.

$f ({\arcsin (x - 2)}^{2}) = \sin (\sqrt{{\arcsin (x - 2)}^{2}}) + 2$

Step 4.3.4

Simplify each term.

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Step 4.3.4.1

Pull terms out from under the radical, assuming positive real numbers.

$f ({\arcsin (x - 2)}^{2}) = \sin (\arcsin (x - 2)) + 2$

Step 4.3.4.2

The functions sine and arcsine are inverses.

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

Step 4.3.5

Combine the opposite terms in $x - 2 + 2$ .

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Step 4.3.5.1

Add $- 2$ and $2$ .

$f ({\arcsin (x - 2)}^{2}) = x + 0$

Step 4.3.5.2

Add $x$ and $0$ .

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

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = {\arcsin (x - 2)}^{2}$ is the inverse of $f (x) = \sin (\sqrt{x}) + 2$ .

$f^{- 1} (x) = {\arcsin (x - 2)}^{2}$