Algebra Examples

$\sin (θ) = - \frac{27}{190}$

Step 1

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

$θ = \arcsin (- \frac{27}{190})$

Step 2

Simplify the right side.

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Step 2.1

Evaluate $\arcsin (- \frac{27}{190})$ .

$θ = - 0.14258793$

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.

$θ = 2 (3.14159265) + 0.14258793 + 3.14159265$

Step 4

Simplify the expression to find the second solution.

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Step 4.1

Subtract $2 π$ from $2 (3.14159265) + 0.14258793 + 3.14159265$ .

$θ = 2 (3.14159265) + 0.14258793 + 3.14159265 - 2 π$

Step 4.2

The resulting angle of $3.28418059$ is positive, less than $2 π$ , and coterminal with $2 (3.14159265) + 0.14258793 + 3.14159265$ .

$θ = 3.28418059$

Step 5

Find the period of $\sin (θ)$ .

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Step 5.1

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

$\frac{2 π}{| b |}$

Step 5.2

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

$\frac{2 π}{| 1 |}$

Step 5.3

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

$\frac{2 π}{1}$

Step 5.4

Divide $2 π$ by $1$ .

$2 π$

Step 6

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

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Step 6.1

Add $2 π$ to $- 0.14258793$ to find the positive angle.

$- 0.14258793 + 2 π$

Step 6.2

Subtract $0.14258793$ from $2 π$ .

$6.14059736$

Step 6.3

List the new angles.

$θ = 6.14059736$

Step 7

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

$θ = 3.28418059 + 2 π n, 6.14059736 + 2 π n$ , for any integer $n$

Step 8

The result can be shown in multiple forms.

Exact Form:

$\sin (θ) = - \frac{27}{190}$

Decimal Form:

$\sin (θ) = - 0.14210526 \dots$