Trigonometry Examples

$\sin (s) = - \frac{\sqrt{3}}{2}$

Step 1

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

$s = \arcsin (- \frac{\sqrt{3}}{2})$

Step 2

Simplify the right side.

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

$s = - \frac{π}{3}$

Step 3

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.

$s = 2 π + \frac{π}{3} + π$

Step 4

Simplify the expression to find the second solution.

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Subtract $2 π$ from $2 π + \frac{π}{3} + π$ .

$s = 2 π + \frac{π}{3} + π - 2 π$

The resulting angle of $\frac{4 π}{3}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{3} + π$ .

$s = \frac{4 π}{3}$

Step 5

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

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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 6

Add $2 π$ to every negative angle to get positive angles.

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Add $2 π$ to $- \frac{π}{3}$ to find the positive angle.

$- \frac{π}{3} + 2 π$

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$2 π \cdot \frac{3}{3} - \frac{π}{3}$

Combine fractions.

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Combine $2 π$ and $\frac{3}{3}$ .

$\frac{2 π \cdot 3}{3} - \frac{π}{3}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

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Multiply $3$ by $2$ .

$\frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$\frac{5 π}{3}$

List the new angles.

$s = \frac{5 π}{3}$

Step 7

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

$s = \frac{4 π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$