Trigonometry Examples

$\sin (θ) > 0$ , $\cos (θ) < 0$

Step 1

Simplify the first inequality.

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Step 1.1

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

$θ > \arcsin (0)$ and $\cos (θ) < 0$

Step 1.2

Simplify the right side.

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Step 1.2.1

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

$θ > 0$ and $\cos (θ) < 0$

Step 1.3

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.

$θ = π - 0$ and $\cos (θ) < 0$

Step 1.4

Subtract $0$ from $π$ .

$θ = π$ and $\cos (θ) < 0$

Step 1.5

Find the period of $\sin (θ)$ .

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Step 1.5.1

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

$\frac{2 π}{| b |}$

Step 1.5.2

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

$\frac{2 π}{| 1 |}$

Step 1.5.3

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

$\frac{2 π}{1}$

Step 1.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.6

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

$θ = 2 π n, π + 2 π n$ and $\cos (θ) < 0$

Step 1.7

Consolidate the answers.

$θ = π n$ and $\cos (θ) < 0$

Step 1.8

Use each root to create test intervals.

$0 < θ < π$

$π < θ < 2 π$ and $\cos (θ) < 0$

Step 1.9

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

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Step 1.9.1

Test a value on the interval $0 < θ < π$ to see if it makes the inequality true.

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Step 1.9.1.1

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

$θ = 2$ and $\cos (θ) < 0$

Step 1.9.1.2

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

$\sin (2) > 0$ and $\cos (θ) < 0$

Step 1.9.1.3

The left side $0.90929742$ is greater than the right side $0$ , which means that the given statement is always true.

True and $\cos (θ) < 0$

Step 1.9.2

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

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Step 1.9.2.1

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

$θ = 5$ and $\cos (θ) < 0$

Step 1.9.2.2

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

$\sin (5) > 0$ and $\cos (θ) < 0$

Step 1.9.2.3

The left side $- 0.95892427$ is not greater than the right side $0$ , which means that the given statement is false.

False and $\cos (θ) < 0$

Step 1.9.3

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

$0 < θ < π$ True

$π < θ < 2 π$ False and $\cos (θ) < 0$

$0 < θ < π$ True

$π < θ < 2 π$ False and $\cos (θ) < 0$

Step 1.10

The solution consists of all of the true intervals.

$0 + 2 π n < θ < π + 2 π n$ and $\cos (θ) < 0$

Step 2

Simplify the second inequality.

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Step 2.1

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

$0 + 2 π n < θ < π + 2 π n$ and $θ < \arccos (0)$

Step 2.2

Simplify the right side.

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Step 2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$0 + 2 π n < θ < π + 2 π n$ and $θ < \frac{π}{2}$

Step 2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$0 + 2 π n < θ < π + 2 π n$ and $θ = 2 π - \frac{π}{2}$

Step 2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 2.4.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$0 + 2 π n < θ < π + 2 π n$ and $θ = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 2.4.2

Combine fractions.

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Step 2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

$0 + 2 π n < θ < π + 2 π n$ and $θ = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 2.4.2.2

Combine the numerators over the common denominator.

$0 + 2 π n < θ < π + 2 π n$ and $θ = \frac{2 π \cdot 2 - π}{2}$

Step 2.4.3

Simplify the numerator.

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Step 2.4.3.1

Multiply $2$ by $2$ .

$0 + 2 π n < θ < π + 2 π n$ and $θ = \frac{4 π - π}{2}$

Step 2.4.3.2

Subtract $π$ from $4 π$ .

$0 + 2 π n < θ < π + 2 π n$ and $θ = \frac{3 π}{2}$

Step 2.5

Find the period of $\cos (θ)$ .

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Step 2.5.1

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

$\frac{2 π}{| b |}$

Step 2.5.2

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

$\frac{2 π}{| 1 |}$

Step 2.5.3

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

$\frac{2 π}{1}$

Step 2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.6

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

$0 + 2 π n < θ < π + 2 π n$ and $θ = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n$

Step 2.7

Consolidate the answers.

$0 + 2 π n < θ < π + 2 π n$ and $θ = \frac{π}{2} + π n$

Step 2.8

Use each root to create test intervals.

$0 + 2 π n < θ < π + 2 π n$ and $\frac{π}{2} < θ < \frac{3 π}{2}$

$\frac{3 π}{2} < θ < \frac{5 π}{2}$

Step 2.9

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

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Step 2.9.1

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

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Step 2.9.1.1

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

$0 + 2 π n < θ < π + 2 π n$ and $θ = 3$

Step 2.9.1.2

Replace $θ$ with $3$ in the original inequality.

$0 + 2 π n < θ < π + 2 π n$ and $\cos (3) < 0$

Step 2.9.1.3

The left side $- 0.98999249$ is less than the right side $0$ , which means that the given statement is always true.

$0 + 2 π n < θ < π + 2 π n$ and True

Step 2.9.2

Test a value on the interval $\frac{3 π}{2} < θ < \frac{5 π}{2}$ to see if it makes the inequality true.

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Step 2.9.2.1

Choose a value on the interval $\frac{3 π}{2} < θ < \frac{5 π}{2}$ and see if this value makes the original inequality true.

$0 + 2 π n < θ < π + 2 π n$ and $θ = 6$

Step 2.9.2.2

Replace $θ$ with $6$ in the original inequality.

$0 + 2 π n < θ < π + 2 π n$ and $\cos (6) < 0$

Step 2.9.2.3

The left side $0.96017028$ is not less than the right side $0$ , which means that the given statement is false.

$0 + 2 π n < θ < π + 2 π n$ and False

Step 2.9.3

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

$0 + 2 π n < θ < π + 2 π n$ and $\frac{π}{2} < θ < \frac{3 π}{2}$ True

$\frac{3 π}{2} < θ < \frac{5 π}{2}$ False

$0 + 2 π n < θ < π + 2 π n$ and $\frac{π}{2} < θ < \frac{3 π}{2}$ True

$\frac{3 π}{2} < θ < \frac{5 π}{2}$ False

Step 2.10

The solution consists of all of the true intervals.

$0 + 2 π n < θ < π + 2 π n$ and $\frac{π}{2} + 2 π n < θ < \frac{3 π}{2} + 2 π n$

Step 3