Trigonometry Examples

$\cos (30 - x)$

Step 1

Interchange the variables.

$x = \cos (30 - y)$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\cos (30 - y) = x$ .

$\cos (30 - y) = x$

Step 2.2

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

$30 - y = \arccos (x)$

Step 2.3

Subtract $30$ from both sides of the equation.

$- y = \arccos (x) - 30$

Step 2.4

Divide each term in $- y = \arccos (x) - 30$ by $- 1$ and simplify.

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Step 2.4.1

Divide each term in $- y = \arccos (x) - 30$ by $- 1$ .

$\frac{- y}{- 1} = \frac{\arccos (x)}{- 1} + \frac{- 30}{- 1}$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{\arccos (x)}{- 1} + \frac{- 30}{- 1}$

Step 2.4.2.2

Divide $y$ by $1$ .

$y = \frac{\arccos (x)}{- 1} + \frac{- 30}{- 1}$

Step 2.4.3

Simplify the right side.

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Step 2.4.3.1

Simplify each term.

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Step 2.4.3.1.1

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

$y = - 1 \cdot \arccos (x) + \frac{- 30}{- 1}$

Step 2.4.3.1.2

Rewrite $- 1 \cdot \arccos (x)$ as $- \arccos (x)$ .

$y = - \arccos (x) + \frac{- 30}{- 1}$

Step 2.4.3.1.3

Divide $- 30$ by $- 1$ .

$y = - \arccos (x) + 30$

Step 3

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

$f^{- 1} (x) = - \arccos (x) + 30$

Step 4

Verify if $f^{- 1} (x) = - \arccos (x) + 30$ is the inverse of $f (x) = \cos (30 - x)$ .

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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 $f^{- 1} (\cos (30 - x))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\cos (30 - x)) = - \arccos (\cos (30 - x)) + 30$

Step 4.2.3

Reorder $30$ and $- x$ .

$f^{- 1} (\cos (30 - x)) = - \arccos (\cos (- x + 30)) + 30$

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 $f (- \arccos (x) + 30)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \arccos (x) + 30) = \cos (30 - (- \arccos (x) + 30))$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Apply the distributive property.

$f (- \arccos (x) + 30) = \cos (30 + \arccos (x) - 1 \cdot 30)$

Step 4.3.3.2

Multiply $- - \arccos (x)$ .

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Step 4.3.3.2.1

Multiply $- 1$ by $- 1$ .

$f (- \arccos (x) + 30) = \cos (30 + 1 \arccos (x) - 1 \cdot 30)$

Step 4.3.3.2.2

Multiply $\arccos (x)$ by $1$ .

$f (- \arccos (x) + 30) = \cos (30 + \arccos (x) - 1 \cdot 30)$

Step 4.3.3.3

Multiply $- 1$ by $30$ .

$f (- \arccos (x) + 30) = \cos (30 + \arccos (x) - 30)$

Step 4.3.4

Combine the opposite terms in $30 + \arccos (x) - 30$ .

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Step 4.3.4.1

Subtract $30$ from $30$ .

$f (- \arccos (x) + 30) = \cos (0 + \arccos (x))$

Step 4.3.4.2

Add $0$ and $\arccos (x)$ .

$f (- \arccos (x) + 30) = \cos (\arccos (x))$

Step 4.3.5

The functions cosine and arccosine are inverses.

$f (- \arccos (x) + 30) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \arccos (x) + 30$ is the inverse of $f (x) = \cos (30 - x)$ .

$f^{- 1} (x) = - \arccos (x) + 30$