Precalculus Examples

$\sin (θ) = \frac{1}{2}$

Step 1

Multiply each term by a factor of $1$ that will equate all the denominators. In this case, all terms need a denominator of $2$ .

$\sin (θ) \cdot \frac{2}{2} = \frac{1}{2}$

Step 2

Multiply the expression by a factor of $1$ to create the least common denominator (LCD) of $2$ .

$\sin (θ) \cdot 2$

Step 3

Move $2$ to the left of $\sin (θ)$ .

$\frac{2 \sin (θ)}{2} = \frac{1}{2}$

Step 4

Simplify $\sin (θ) \cdot \frac{2}{2}$ .

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Divide $2$ by $2$ .

$\sin (θ) \cdot 1 = \frac{1}{2}$

Multiply $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{1}{2}$

Step 5

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

$θ = \arcsin (\frac{1}{2})$

Step 6

Simplify the right side.

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The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$θ = \frac{π}{6}$

Step 7

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.

$θ = π - \frac{π}{6}$

Step 8

Simplify $π - \frac{π}{6}$ .

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

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

Combine fractions.

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Combine $π$ and $\frac{6}{6}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

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Move $6$ to the left of $π$ .

$θ = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{6}$

Step 9

Find the period of $\sin (θ)$ .

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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 $1$ .

$2 π$

Step 10

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

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