Solve for a in Degrees sin(a)=0.5

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Trigonometry

$\sin (a) = 0.5$

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

$a = \arcsin (0.5)$

Simplify the right side.

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Evaluate $\arcsin (0.5)$ .

$a = 30$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the second quadrant.

$a = 180 - 30$

Subtract $30$ from $180$ .

$a = 150$

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

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

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

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

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

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

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

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

$a = 30 + 360 n, 150 + 360 n$ , for any integer $n$

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