Precalculus Examples

$\sin (2 θ) = - \frac{1}{2}$

Step 1

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

$2 θ = \arcsin (- \frac{1}{2})$

Step 2

Simplify the right side.

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Step 2.1

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

$2 θ = - \frac{π}{6}$

Step 3

Divide each term in $2 θ = - \frac{π}{6}$ by $2$ and simplify.

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Step 3.1

Divide each term in $2 θ = - \frac{π}{6}$ by $2$ .

$\frac{2 θ}{2} = \frac{- \frac{π}{6}}{2}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Cancel the common factor of $2$ .

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Step 3.2.1.1

Cancel the common factor.

$\frac{2 θ}{2} = \frac{- \frac{π}{6}}{2}$

Step 3.2.1.2

Divide $θ$ by $1$ .

$θ = \frac{- \frac{π}{6}}{2}$

Step 3.3

Simplify the right side.

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Step 3.3.1

Multiply the numerator by the reciprocal of the denominator.

$θ = - \frac{π}{6} \cdot \frac{1}{2}$

Step 3.3.2

Multiply $- \frac{π}{6} \cdot \frac{1}{2}$ .

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Step 3.3.2.1

Multiply $\frac{1}{2}$ by $\frac{π}{6}$ .

$θ = - \frac{π}{2 \cdot 6}$

Step 3.3.2.2

Multiply $2$ by $6$ .

$θ = - \frac{π}{12}$

Step 4

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$2 θ = 2 π + \frac{π}{6} + π$

Step 5

Simplify the expression to find the second solution.

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Step 5.1

Subtract $2 π$ from $2 π + \frac{π}{6} + π$ .

$2 θ = 2 π + \frac{π}{6} + π - 2 π$

Step 5.2

The resulting angle of $\frac{7 π}{6}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{6} + π$ .

$2 θ = \frac{7 π}{6}$

Step 5.3

Divide each term in $2 θ = \frac{7 π}{6}$ by $2$ and simplify.

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Step 5.3.1

Divide each term in $2 θ = \frac{7 π}{6}$ by $2$ .

$\frac{2 θ}{2} = \frac{\frac{7 π}{6}}{2}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $2$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{2 θ}{2} = \frac{\frac{7 π}{6}}{2}$

Step 5.3.2.1.2

Divide $θ$ by $1$ .

$θ = \frac{\frac{7 π}{6}}{2}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{7 π}{6} \cdot \frac{1}{2}$

Step 5.3.3.2

Multiply $\frac{7 π}{6} \cdot \frac{1}{2}$ .

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Step 5.3.3.2.1

Multiply $\frac{7 π}{6}$ by $\frac{1}{2}$ .

$θ = \frac{7 π}{6 \cdot 2}$

Step 5.3.3.2.2

Multiply $6$ by $2$ .

$θ = \frac{7 π}{12}$

Step 6

Find the period of $\sin (2 θ)$ .

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Step 6.1

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

$\frac{2 π}{| b |}$

Step 6.2

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Step 6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Step 6.4

Cancel the common factor of $2$ .

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Step 6.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 6.4.2

Divide $π$ by $1$ .

$π$

Step 7

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

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Step 7.1

Add $π$ to $- \frac{π}{12}$ to find the positive angle.

$- \frac{π}{12} + π$

Step 7.2

To write $π$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$π \cdot \frac{12}{12} - \frac{π}{12}$

Step 7.3

Combine fractions.

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Step 7.3.1

Combine $π$ and $\frac{12}{12}$ .

$\frac{π \cdot 12}{12} - \frac{π}{12}$

Step 7.3.2

Combine the numerators over the common denominator.

$\frac{π \cdot 12 - π}{12}$

Step 7.4

Simplify the numerator.

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Step 7.4.1

Move $12$ to the left of $π$ .

$\frac{12 \cdot π - π}{12}$

Step 7.4.2

Subtract $π$ from $12 π$ .

$\frac{11 π}{12}$

Step 7.5

List the new angles.

$θ = \frac{11 π}{12}$

Step 8

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

$θ = \frac{7 π}{12} + π n, \frac{11 π}{12} + π n$ , for any integer $n$