Trigonometry Examples

$5 \cos (5 x) + 2$

Step 1

Interchange the variables.

$x = 5 \cos (5 y) + 2$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $5 \cos (5 y) + 2 = x$ .

$5 \cos (5 y) + 2 = x$

Step 2.2

Subtract $2$ from both sides of the equation.

$5 \cos (5 y) = x - 2$

Step 2.3

Divide each term in $5 \cos (5 y) = x - 2$ by $5$ and simplify.

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Step 2.3.1

Divide each term in $5 \cos (5 y) = x - 2$ by $5$ .

$\frac{5 \cos (5 y)}{5} = \frac{x}{5} + \frac{- 2}{5}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $5$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{5 \cos (5 y)}{5} = \frac{x}{5} + \frac{- 2}{5}$

Step 2.3.2.1.2

Divide $\cos (5 y)$ by $1$ .

$\cos (5 y) = \frac{x}{5} + \frac{- 2}{5}$

Step 2.3.3

Simplify the right side.

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Step 2.3.3.1

Move the negative in front of the fraction.

$\cos (5 y) = \frac{x}{5} - \frac{2}{5}$

Step 2.4

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

$5 y = \arccos (\frac{x}{5} - \frac{2}{5})$

Step 2.5

Divide each term in $5 y = \arccos (\frac{x}{5} - \frac{2}{5})$ by $5$ and simplify.

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Step 2.5.1

Divide each term in $5 y = \arccos (\frac{x}{5} - \frac{2}{5})$ by $5$ .

$\frac{5 y}{5} = \frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}$

Step 2.5.2

Simplify the left side.

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Step 2.5.2.1

Cancel the common factor of $5$ .

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Step 2.5.2.1.1

Cancel the common factor.

$\frac{5 y}{5} = \frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}$

Step 2.5.2.1.2

Divide $y$ by $1$ .

$y = \frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}$

Step 3

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

$f^{- 1} (x) = \frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}$

Step 4

Verify if $f^{- 1} (x) = \frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}$ is the inverse of $f (x) = 5 \cos (5 x) + 2$ .

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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 $f^{- 1} (5 \cos (5 x) + 2)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (5 \cos (5 x) + 2) = \frac{\arccos (\frac{5 \cos (5 x) + 2}{5} - \frac{2}{5})}{5}$

Step 4.2.3

Simplify the numerator.

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Step 4.2.3.1

Combine the numerators over the common denominator.

$f^{- 1} (5 \cos (5 x) + 2) = \frac{\arccos (\frac{5 \cos (5 x) + 2 - 2}{5})}{5}$

Step 4.2.3.2

Combine the opposite terms in $5 \cos (5 x) + 2 - 2$ .

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Step 4.2.3.2.1

Subtract $2$ from $2$ .

$f^{- 1} (5 \cos (5 x) + 2) = \frac{\arccos (\frac{5 \cos (5 x) + 0}{5})}{5}$

Step 4.2.3.2.2

Add $5 \cos (5 x)$ and $0$ .

$f^{- 1} (5 \cos (5 x) + 2) = \frac{\arccos (\frac{5 \cos (5 x)}{5})}{5}$

Step 4.2.3.3

Cancel the common factor of $5$ .

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Step 4.2.3.3.1

Cancel the common factor.

$f^{- 1} (5 \cos (5 x) + 2) = \frac{\arccos (\frac{5 \cos (5 x)}{5})}{5}$

Step 4.2.3.3.2

Divide $\cos (5 x)$ by $1$ .

$f^{- 1} (5 \cos (5 x) + 2) = \frac{\arccos (\cos (5 x))}{5}$

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 $f (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}) = 5 \cos (5 (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5})) + 2$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Combine the numerators over the common denominator.

$f (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}) = 5 \cos (5 (\frac{\arccos (\frac{x - 2}{5})}{5})) + 2$

Step 4.3.3.2

Cancel the common factor of $5$ .

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Step 4.3.3.2.1

Cancel the common factor.

$f (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}) = 5 \cos (5 (\frac{\arccos (\frac{x - 2}{5})}{5})) + 2$

Step 4.3.3.2.2

Rewrite the expression.

$f (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}) = 5 \cos (\arccos (\frac{x - 2}{5})) + 2$

Step 4.3.3.3

The functions cosine and arccosine are inverses.

$f (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}) = 5 (\frac{x - 2}{5}) + 2$

Step 4.3.3.4

Cancel the common factor of $5$ .

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Step 4.3.3.4.1

Cancel the common factor.

$f (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}) = 5 (\frac{x - 2}{5}) + 2$

Step 4.3.3.4.2

Rewrite the expression.

$f (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}) = x - 2 + 2$

Step 4.3.4

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

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Step 4.3.4.1

Add $- 2$ and $2$ .

$f (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}) = x + 0$

Step 4.3.4.2

Add $x$ and $0$ .

$f (\frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}) = x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}$ is the inverse of $f (x) = 5 \cos (5 x) + 2$ .

$f^{- 1} (x) = \frac{\arccos (\frac{x}{5} - \frac{2}{5})}{5}$