Trigonometry Examples

$\ln (\arctan (6 x^{2}))$

Step 1

Find the asymptotes.

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Step 1.1

Set the argument of the logarithm equal to zero.

$\arctan (6 x^{2}) = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

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

$6 x^{2} = \tan (0)$

Step 1.2.2

Simplify the right side.

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Step 1.2.2.1

The exact value of $\tan (0)$ is $0$ .

$6 x^{2} = 0$

Step 1.2.3

Divide each term in $6 x^{2} = 0$ by $6$ and simplify.

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Step 1.2.3.1

Divide each term in $6 x^{2} = 0$ by $6$ .

$\frac{6 x^{2}}{6} = \frac{0}{6}$

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Cancel the common factor of $6$ .

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Step 1.2.3.2.1.1

Cancel the common factor.

$\frac{6 x^{2}}{6} = \frac{0}{6}$

Step 1.2.3.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{0}{6}$

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Divide $0$ by $6$ .

$x^{2} = 0$

Step 1.2.4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Step 1.2.5

Simplify $\pm \sqrt{0}$ .

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Step 1.2.5.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 1.2.5.2

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

$x = \pm 0$

Step 1.2.5.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 1.3

The vertical asymptote occurs at $x = 0$ .

Vertical Asymptote: $x = 0$

Step 2

Find the point at $x = 1$ .

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Step 2.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \ln (\arctan (6 {(1)}^{2}))$

Step 2.2

Simplify the result.

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Step 2.2.1

One to any power is one.

$f (1) = \ln (\arctan (6 \cdot 1))$

Step 2.2.2

Multiply $6$ by $1$ .

$f (1) = \ln (\arctan (6))$

Step 2.2.3

Evaluate $\arctan (6)$ .

$f (1) = \ln (1.40564764)$

Step 2.2.4

The final answer is $\ln (1.40564764)$ .

$\ln (1.40564764)$

Step 3

Find the point at $x = 2$ .

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Step 3.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = \ln (\arctan (6 {(2)}^{2}))$

Step 3.2

Simplify the result.

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Step 3.2.1

Raise $2$ to the power of $2$ .

$f (2) = \ln (\arctan (6 \cdot 4))$

Step 3.2.2

Multiply $6$ by $4$ .

$f (2) = \ln (\arctan (24))$

Step 3.2.3

Evaluate $\arctan (24)$ .

$f (2) = \ln (1.52915374)$

Step 3.2.4

The final answer is $\ln (1.52915374)$ .

$\ln (1.52915374)$

Step 4

Find the point at $x = 3$ .

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Step 4.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = \ln (\arctan (6 {(3)}^{2}))$

Step 4.2

Simplify the result.

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Step 4.2.1

Raise $3$ to the power of $2$ .

$f (3) = \ln (\arctan (6 \cdot 9))$

Step 4.2.2

Multiply $6$ by $9$ .

$f (3) = \ln (\arctan (54))$

Step 4.2.3

Evaluate $\arctan (54)$ .

$f (3) = \ln (1.55227992)$

Step 4.2.4

The final answer is $\ln (1.55227992)$ .

$\ln (1.55227992)$

Step 5

The log function can be graphed using the vertical asymptote at $x = 0$ and the points $(1, 0.34049815), (2, 0.42471447), (3, 0.43972476)$ .

Vertical Asymptote: $x = 0$

$\begin{array}{c} x & y \\ 1 & 0.34 \\ 2 & 0.425 \\ 3 & 0.44 \end{array}$

Step 6