Trigonometry Examples

$\sin (θ) = - \frac{\sqrt{2}}{2}$

Step 1

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

$θ = \arcsin (- \frac{\sqrt{2}}{2})$

Step 2

Simplify the right side.

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Step 2.1

The exact value of $\arcsin (- \frac{\sqrt{2}}{2})$ is $- 45$ .

$θ = - 45$

Step 3

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $360$ , to find a reference angle. Next, add this reference angle to $180$ to find the solution in the third quadrant.

$θ = 360 + 45 + 180$

Step 4

Simplify the expression to find the second solution.

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Step 4.1

Subtract $360 °$ from $360 + 45 + 180 °$ .

$θ = 360 + 45 + 180 ° - 360 °$

Step 4.2

The resulting angle of $225 °$ is positive, less than $360 °$ , and coterminal with $360 + 45 + 180$ .

$θ = 225 °$

Step 5

Find the period of $\sin (θ)$ .

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Step 5.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 5.2

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

$\frac{360}{| 1 |}$

Step 5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Step 5.4

Divide $360$ by $1$ .

$360$

Step 6

Add $360$ to every negative angle to get positive angles.

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Step 6.1

Add $360$ to $- 45$ to find the positive angle.

$- 45 + 360$

Step 6.2

Subtract $45$ from $360$ .

$315$

Step 6.3

List the new angles.

$θ = 315$

Step 7

The period of the $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 225 + 360 n, 315 + 360 n$ , for any integer $n$