Trigonometry Examples

$y = 1 + 2 \arctan (x)$

Step 1

Interchange the variables.

$x = 1 + 2 \arctan (y)$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $1 + 2 \arctan (y) = x$ .

$1 + 2 \arctan (y) = x$

Step 2.2

Subtract $1$ from both sides of the equation.

$2 \arctan (y) = x - 1$

Step 2.3

Divide each term in $2 \arctan (y) = x - 1$ by $2$ and simplify.

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Step 2.3.1

Divide each term in $2 \arctan (y) = x - 1$ by $2$ .

$\frac{2 \arctan (y)}{2} = \frac{x}{2} + \frac{- 1}{2}$

Step 2.3.2

Simplify the left side.

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Step 2.3.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.1.1

Cancel the common factor.

$\frac{2 \arctan (y)}{2} = \frac{x}{2} + \frac{- 1}{2}$

Step 2.3.2.1.2

Divide $\arctan (y)$ by $1$ .

$\arctan (y) = \frac{x}{2} + \frac{- 1}{2}$

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.

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

Step 2.4

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

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

Step 3

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

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

Step 4

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

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

$f^{- 1} (1 + 2 \arctan (x)) = \tan (\frac{1 + 2 \arctan (x)}{2} - \frac{1}{2})$

Step 4.2.3

Combine the numerators over the common denominator.

$f^{- 1} (1 + 2 \arctan (x)) = \tan (\frac{1 + 2 \arctan (x) - 1}{2})$

Step 4.2.4

Combine the opposite terms in $1 + 2 \arctan (x) - 1$ .

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Step 4.2.4.1

Subtract $1$ from $1$ .

$f^{- 1} (1 + 2 \arctan (x)) = \tan (\frac{2 \arctan (x) + 0}{2})$

Step 4.2.4.2

Add $2 \arctan (x)$ and $0$ .

$f^{- 1} (1 + 2 \arctan (x)) = \tan (\frac{2 \arctan (x)}{2})$

Step 4.2.5

Cancel the common factor of $2$ .

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Step 4.2.5.1

Cancel the common factor.

$f^{- 1} (1 + 2 \arctan (x)) = \tan (\frac{2 \arctan (x)}{2})$

Step 4.2.5.2

Divide $\arctan (x)$ by $1$ .

$f^{- 1} (1 + 2 \arctan (x)) = \tan (\arctan (x))$

Step 4.2.6

The functions tangent and arctangent are inverses.

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

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

$f (\tan (\frac{x}{2} - \frac{1}{2})) = 1 + 2 \arctan (\tan (\frac{x}{2} - \frac{1}{2}))$

Step 4.4

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

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