Trigonometry Examples

$\sin (x) > 0.5$

Step 1

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

$x > \arcsin (0.5)$

Step 2

Simplify the right side.

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Step 2.1

Evaluate $\arcsin (0.5)$ .

$x > \frac{π}{6}$

Step 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.

$x = π - \frac{π}{6}$

Step 4

Simplify $π - \frac{π}{6}$ .

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Step 4.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 4.2

Combine fractions.

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Step 4.2.1

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 4.3

Simplify the numerator.

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Step 4.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 4.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 5

Find the period of $\sin (x)$ .

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Step 5.1

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

$\frac{2 π}{| b |}$

Step 5.2

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

$\frac{2 π}{| 1 |}$

Step 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 5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6

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

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 7

Use each root to create test intervals.

$\frac{π}{6} < x < \frac{5 π}{6}$

$\frac{5 π}{6} < x < \frac{13 π}{6}$

$\frac{13 π}{6} < x < \frac{17 π}{6}$

Step 8

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 8.1

Test a value on the interval $\frac{π}{6} < x < \frac{5 π}{6}$ to see if it makes the inequality true.

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Step 8.1.1

Choose a value on the interval $\frac{π}{6} < x < \frac{5 π}{6}$ and see if this value makes the original inequality true.

$x = 2$

Step 8.1.2

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

$\sin (2) > 0.5$

Step 8.1.3

The left side $0.90929742$ is greater than the right side $0.5$ , which means that the given statement is always true.

True

Step 8.2

Test a value on the interval $\frac{5 π}{6} < x < \frac{13 π}{6}$ to see if it makes the inequality true.

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Step 8.2.1

Choose a value on the interval $\frac{5 π}{6} < x < \frac{13 π}{6}$ and see if this value makes the original inequality true.

$x = 5$

Step 8.2.2

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

$\sin (5) > 0.5$

Step 8.2.3

The left side $- 0.95892427$ is not greater than the right side $0.5$ , which means that the given statement is false.

False

Step 8.3

Test a value on the interval $\frac{13 π}{6} < x < \frac{17 π}{6}$ to see if it makes the inequality true.

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Step 8.3.1

Choose a value on the interval $\frac{13 π}{6} < x < \frac{17 π}{6}$ and see if this value makes the original inequality true.

$x = 8$

Step 8.3.2

Replace $x$ with $8$ in the original inequality.

$\sin (8) > 0.5$

Step 8.3.3

The left side $0.98935824$ is greater than the right side $0.5$ , which means that the given statement is always true.

True

Step 8.4

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

$\frac{π}{6} < x < \frac{5 π}{6}$ True

$\frac{5 π}{6} < x < \frac{13 π}{6}$ False

$\frac{13 π}{6} < x < \frac{17 π}{6}$ True

$\frac{π}{6} < x < \frac{5 π}{6}$ True

$\frac{5 π}{6} < x < \frac{13 π}{6}$ False

$\frac{13 π}{6} < x < \frac{17 π}{6}$ True

Step 9

The solution consists of all of the true intervals.

$\frac{π}{6} + 2 π n < x < \frac{5 π}{6} + 2 π n$ or $\frac{13 π}{6} + 2 π n < x < \frac{17 π}{6} + 2 π n$ , for any integer $n$

Step 10