Graph tan(theta)>0 , sin(theta)>0

$\tan (θ) > 0$ , $\sin (θ) > 0$

Solve for $θ$ .

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

$θ > \arctan (0)$

Simplify the right side.

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

$θ > 0$

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

$θ = π + 0$

Add $π$ and $0$ .

$θ = π$

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 $1$ .

$π$

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

$θ = π n, π + π n$ , for any integer $n$

Consolidate the answers.

$θ = π n$ , for any integer $n$

Find the domain of $\tan (θ)$ .

Set the argument in $\tan (θ)$ equal to $\frac{π}{2} + π n$ to find where the expression is undefined.

$θ = \frac{π}{2} + π n$ , for any integer $n$

The domain is all values of $θ$ that make the expression defined.

${θ | θ \neq \frac{π}{2} + π n}$ $n$ , for any integer $n$

Use each root to create test intervals.

$0 < θ < \frac{π}{2}$

$\frac{π}{2} < θ < π$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $0 < θ < \frac{π}{2}$ to see if it makes the inequality true.

Choose a value on the interval $0 < θ < \frac{π}{2}$ and see if this value makes the original inequality true.

$θ = 1$

Replace $θ$ with $1$ in the original inequality.

$\tan (1) > 0$

The left side $1.55740772$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{π}{2} < θ < π$ to see if it makes the inequality true.

Choose a value on the interval $\frac{π}{2} < θ < π$ and see if this value makes the original inequality true.

$θ = 2$

Replace $θ$ with $2$ in the original inequality.

$\tan (2) > 0$

The left side $- 2.18503986$ is not greater than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$0 < θ < \frac{π}{2}$ True

$\frac{π}{2} < θ < π$ False

$0 < θ < \frac{π}{2}$ True

$\frac{π}{2} < θ < π$ False

The solution consists of all of the true intervals.

$0 + π n < θ < \frac{π}{2} + π n$ , for any integer $n$

Solve for $θ$ .

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

$θ > \arcsin (0)$

Simplify the right side.

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

$θ > 0$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$θ = π - 0$

Subtract $0$ from $π$ .

$θ = π$

Find the period of $\sin (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\sin (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = 2 π n, π + 2 π n$ , for any integer $n$

Consolidate the answers.

$θ = π n$ , for any integer $n$

Find the domain of $\tan (θ)$ .

Set the argument in $\tan (θ)$ equal to $\frac{π}{2} + π n$ to find where the expression is undefined.

$θ = \frac{π}{2} + π n$ , for any integer $n$

The domain is all values of $θ$ that make the expression defined.

${θ | θ \neq \frac{π}{2} + π n}$ $n$ , for any integer $n$

Use each root to create test intervals.

$0 < θ < \frac{π}{2}$

$\frac{π}{2} < θ < π$

$π < θ < \frac{3 π}{2}$

$\frac{3 π}{2} < θ < 2 π$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $0 < θ < \frac{π}{2}$ to see if it makes the inequality true.

Choose a value on the interval $0 < θ < \frac{π}{2}$ and see if this value makes the original inequality true.

$θ = 1$

Replace $θ$ with $1$ in the original inequality.

$\tan (1) > 0$

The left side $1.55740772$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{π}{2} < θ < π$ to see if it makes the inequality true.

Choose a value on the interval $\frac{π}{2} < θ < π$ and see if this value makes the original inequality true.

$θ = 2$

Replace $θ$ with $2$ in the original inequality.

$\tan (2) > 0$

The left side $- 2.18503986$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $π < θ < \frac{3 π}{2}$ to see if it makes the inequality true.

Choose a value on the interval $π < θ < \frac{3 π}{2}$ and see if this value makes the original inequality true.

$θ = 4$

Replace $θ$ with $4$ in the original inequality.

$\tan (4) > 0$

The left side $1.15782128$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{3 π}{2} < θ < 2 π$ to see if it makes the inequality true.

Choose a value on the interval $\frac{3 π}{2} < θ < 2 π$ and see if this value makes the original inequality true.

$θ = 5$

Replace $θ$ with $5$ in the original inequality.

$\tan (5) > 0$

The left side $- 3.380515$ is not greater than the right side $0$ , which means that the given statement is false.

False

Compare the intervals to determine which ones satisfy the original inequality.

$0 < θ < \frac{π}{2}$ True

$\frac{π}{2} < θ < π$ False

$π < θ < \frac{3 π}{2}$ True

$\frac{3 π}{2} < θ < 2 π$ False

$0 < θ < \frac{π}{2}$ True

$\frac{π}{2} < θ < π$ False

$π < θ < \frac{3 π}{2}$ True

$\frac{3 π}{2} < θ < 2 π$ False

The solution consists of all of the true intervals.

$0 + 2 π n < θ < \frac{π}{2} + 2 π n$ or $π + 2 π n < θ < \frac{3 π}{2} + 2 π n$ , for any integer $n$

Combine the intervals.

$2 π n < θ < \frac{π}{2} + 2 π n or π + 2 π n < θ < \frac{3 π}{2} + 2 π n$ , for any integer $n$

Plot each graph on the same coordinate system.

$\tan (θ) > 0$

$\sin (θ) > 0$

Graph tan(theta)>0 , sin(theta)>0

Related Questions

Support LCMGCF.com by disabling your adblocker.