Evaluate sin(2theta)=( square root of 3)/2

$\sin (2 θ) = \frac{\sqrt{3}}{2}$

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

$2 θ = \arcsin (\frac{\sqrt{3}}{2})$

Simplify the right side.

The exact value of $\arcsin (\frac{\sqrt{3}}{2})$ is $\frac{π}{3}$ .

$2 θ = \frac{π}{3}$

Divide each term in $2 θ = \frac{π}{3}$ by $2$ and simplify.

Divide each term in $2 θ = \frac{π}{3}$ by $2$ .

$\frac{2 θ}{2} = \frac{\frac{π}{3}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 θ}{2} = \frac{\frac{π}{3}}{2}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{π}{3}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{π}{3} \cdot \frac{1}{2}$

Multiply $\frac{π}{3} \cdot \frac{1}{2}$ .

Multiply $\frac{π}{3}$ by $\frac{1}{2}$ .

$θ = \frac{π}{3 \cdot 2}$

Multiply $3$ by $2$ .

$θ = \frac{π}{6}$

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.

$2 θ = π - \frac{π}{3}$

Solve for $θ$ .

Simplify.

To write $π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$2 θ = π \cdot \frac{3}{3} - \frac{π}{3}$

Combine $π$ and $\frac{3}{3}$ .

$2 θ = \frac{π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$2 θ = \frac{π \cdot 3 - π}{3}$

Subtract $π$ from $π \cdot 3$ .

Reorder $π$ and $3$ .

$2 θ = \frac{3 \cdot π - π}{3}$

Subtract $π$ from $3 \cdot π$ .

$2 θ = \frac{2 \cdot π}{3}$

Divide each term in $2 θ = \frac{2 \cdot π}{3}$ by $2$ and simplify.

Divide each term in $2 θ = \frac{2 \cdot π}{3}$ by $2$ .

$\frac{2 θ}{2} = \frac{\frac{2 \cdot π}{3}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 θ}{2} = \frac{\frac{2 \cdot π}{3}}{2}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{2 \cdot π}{3}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{2 \cdot π}{3} \cdot \frac{1}{2}$

Cancel the common factor of $2$ .

Factor $2$ out of $2 \cdot π$ .

$θ = \frac{2 (π)}{3} \cdot \frac{1}{2}$

Cancel the common factor.

$θ = \frac{2 π}{3} \cdot \frac{1}{2}$

Rewrite the expression.

$θ = \frac{π}{3}$

Find the period of $\sin (2 θ)$ .

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 2 |}$

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

$\frac{2 π}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 π}{2}$

Divide $π$ by $1$ .

$π$

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

$θ = \frac{π}{6} + π n, \frac{π}{3} + π n$ , for any integer $n$

Evaluate sin(2theta)=( square root of 3)/2

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