Evaluate tan(theta)=-1

$\tan (θ) = - 1$

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

$θ = \arctan (- 1)$

Simplify the right side.

The exact value of $\arctan (- 1)$ is $- \frac{π}{4}$ .

$θ = - \frac{π}{4}$

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

$θ = - \frac{π}{4} - π$

Simplify the expression to find the second solution.

Add $2 π$ to $- \frac{π}{4} - π$ .

$θ = - \frac{π}{4} - π + 2 π$

The resulting angle of $\frac{3 π}{4}$ is positive and coterminal with $- \frac{π}{4} - π$ .

$θ = \frac{3 π}{4}$

Find the period of $\tan (θ)$ .

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

Add $π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + π$

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$π \cdot \frac{4}{4} - \frac{π}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$\frac{π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$\frac{4 \cdot π - π}{4}$

Subtract $π$ from $4 π$ .

$\frac{3 π}{4}$

List the new angles.

$θ = \frac{3 π}{4}$

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

$θ = \frac{3 π}{4} + π n, \frac{3 π}{4} + π n$ , for any integer $n$

Consolidate the answers.

$θ = \frac{3 π}{4} + π n$ , for any integer $n$

Evaluate tan(theta)=-1

Related Questions

Support LCMGCF.com by disabling your adblocker.