Trigonometry Examples

$\tan (θ) = \frac{- \sqrt{3}}{3}$

Step 1

Move the negative in front of the fraction.

$\tan (θ) = - \frac{\sqrt{3}}{3}$

Step 2

Take the inverse tangent of both sides of the equation to extract $θ$ from inside the tangent.

$θ = \arctan (- \frac{\sqrt{3}}{3})$

Step 3

Simplify the right side.

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Step 3.1

The exact value of $\arctan (- \frac{\sqrt{3}}{3})$ is $- 30$ .

$θ = - 30$

Step 4

The tangent function is negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the third quadrant.

$θ = - 30 - 180$

Step 5

Simplify the expression to find the second solution.

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Step 5.1

Add $360 °$ to $- 30 - 180 °$ .

$θ = - 30 - 180 ° + 360 °$

Step 5.2

The resulting angle of $150 °$ is positive and coterminal with $- 30 - 180$ .

$θ = 150 °$

Step 6

Find the period of $\tan (θ)$ .

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Step 6.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

Step 6.2

Replace $b$ with $1$ in the formula for period.

$\frac{180}{| 1 |}$

Step 6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{180}{1}$

Step 6.4

Divide $180$ by $1$ .

$180$

Step 7

Add $180$ to every negative angle to get positive angles.

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Step 7.1

Add $180$ to $- 30$ to find the positive angle.

$- 30 + 180$

Step 7.2

Subtract $30$ from $180$ .

$150$

Step 7.3

List the new angles.

$θ = 150$

Step 8

The period of the $\tan (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 150 + 180 n, 150 + 180 n$ , for any integer $n$

Step 9

Consolidate the answers.

$θ = 150 + 180 n$ , for any integer $n$