Trigonometry Examples

$g (x) = \sin (x + \frac{π}{3}) - 7$

Step 1

Write $g (x) = \sin (x + \frac{π}{3}) - 7$ as an equation.

$y = \sin (x + \frac{π}{3}) - 7$

Step 2

Interchange the variables.

$x = \sin (y + \frac{π}{3}) - 7$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\sin (y + \frac{π}{3}) - 7 = x$ .

$\sin (y + \frac{π}{3}) - 7 = x$

Step 3.2

Add $7$ to both sides of the equation.

$\sin (y + \frac{π}{3}) = x + 7$

Step 3.3

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

$y + \frac{π}{3} = \arcsin (x + 7)$

Step 3.4

Subtract $\frac{π}{3}$ from both sides of the equation.

$y = \arcsin (x + 7) - \frac{π}{3}$

Step 4

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

$g^{- 1} (x) = \arcsin (x + 7) - \frac{π}{3}$

Step 5

Verify if $g^{- 1} (x) = \arcsin (x + 7) - \frac{π}{3}$ is the inverse of $g (x) = \sin (x + \frac{π}{3}) - 7$ .

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

Evaluate $g^{- 1} (\sin (x + \frac{π}{3}) - 7)$ by substituting in the value of $g$ into $g^{- 1}$ .

$g^{- 1} (\sin (x + \frac{π}{3}) - 7) = \arcsin ((\sin (x + \frac{π}{3}) - 7) + 7) - \frac{π}{3}$

Step 5.2.3

Combine the opposite terms in $\arcsin ((\sin (x + \frac{π}{3}) - 7) + 7) - \frac{π}{3}$ .

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Step 5.2.3.1

Add $- 7$ and $7$ .

$g^{- 1} (\sin (x + \frac{π}{3}) - 7) = \arcsin (\sin (x + \frac{π}{3}) + 0) - \frac{π}{3}$

Step 5.2.3.2

Add $\sin (x + \frac{π}{3})$ and $0$ .

$g^{- 1} (\sin (x + \frac{π}{3}) - 7) = \arcsin (\sin (x + \frac{π}{3})) - \frac{π}{3}$

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate $g (\arcsin (x + 7) - \frac{π}{3})$ by substituting in the value of $g^{- 1}$ into $g$ .

$g (\arcsin (x + 7) - \frac{π}{3}) = \sin ((\arcsin (x + 7) - \frac{π}{3}) + \frac{π}{3}) - 7$

Step 5.3.3

Combine the opposite terms in $\sin ((\arcsin (x + 7) - \frac{π}{3}) + \frac{π}{3}) - 7$ .

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Step 5.3.3.1

Add $- \frac{π}{3}$ and $\frac{π}{3}$ .

$g (\arcsin (x + 7) - \frac{π}{3}) = \sin (\arcsin (x + 7) + 0) - 7$

Step 5.3.3.2

Add $\arcsin (x + 7)$ and $0$ .

$g (\arcsin (x + 7) - \frac{π}{3}) = \sin (\arcsin (x + 7)) - 7$

Step 5.3.4

The functions sine and arcsine are inverses.

$g (\arcsin (x + 7) - \frac{π}{3}) = x + 7 - 7$

Step 5.3.5

Combine the opposite terms in $x + 7 - 7$ .

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Step 5.3.5.1

Subtract $7$ from $7$ .

$g (\arcsin (x + 7) - \frac{π}{3}) = x + 0$

Step 5.3.5.2

Add $x$ and $0$ .

$g (\arcsin (x + 7) - \frac{π}{3}) = x$

Step 5.4

Since $g^{- 1} (g (x)) = x$ and $g (g^{- 1} (x)) = x$ , then $g^{- 1} (x) = \arcsin (x + 7) - \frac{π}{3}$ is the inverse of $g (x) = \sin (x + \frac{π}{3}) - 7$ .

$g^{- 1} (x) = \arcsin (x + 7) - \frac{π}{3}$