Given that [tex]\(\cot (\theta) = -\frac{1}{2}\)[/tex] and [tex]\(\theta\)[/tex] is in Quadrant II, we want to find [tex]\(\sin (\theta)\)[/tex].
1. Express [tex]\(\cot (\theta)\)[/tex] as [tex]\(\frac{\cos (\theta)}{\sin (\theta)}\)[/tex]:
[tex]\[
\cot (\theta) = \frac{\cos (\theta)}{\sin (\theta)} = -\frac{1}{2}
\][/tex]
2. Express [tex]\(\cos (\theta)\)[/tex] in terms of [tex]\(\sin (\theta)\)[/tex]:
[tex]\[
\cos (\theta) = -\frac{1}{2} \sin (\theta)
\][/tex]
3. Use the Pythagorean identity:
[tex]\[
\sin^2 (\theta) + \cos^2 (\theta) = 1
\][/tex]
4. Substitute [tex]\(\cos (\theta) = -\frac{1}{2} \sin (\theta)\)[/tex] into the Pythagorean identity:
[tex]\[
\sin^2 (\theta) + \left( -\frac{1}{2} \sin (\theta) \right)^2 = 1
\][/tex]
5. Simplify the equation:
[tex]\[
\sin^2 (\theta) + \frac{1}{4} \sin^2 (\theta) = 1
\][/tex]
6. Combine like terms:
[tex]\[
\sin^2 (\theta) \left( 1 + \frac{1}{4} \right) = 1
\][/tex]
[tex]\[
\sin^2 (\theta) \left( \frac{4}{4} + \frac{1}{4} \right) = 1
\][/tex]
[tex]\[
\sin^2 (\theta) \left( \frac{5}{4} \right) = 1
\][/tex]
7. Solve for [tex]\(\sin^2 (\theta)\)[/tex]:
[tex]\[
\sin^2 (\theta) = \frac{4}{5}
\][/tex]
8. Take the square root of both sides to solve for [tex]\(\sin (\theta)\)[/tex]:
[tex]\[
\sin (\theta) = \pm \sqrt{\frac{4}{5}}
\][/tex]
[tex]\[
\sin (\theta) = \pm \frac{2}{\sqrt{5}}
\][/tex]
9. Since [tex]\(\theta\)[/tex] is in Quadrant II where [tex]\(\sin (\theta)\)[/tex] is positive:
[tex]\[
\sin (\theta) = \frac{2}{\sqrt{5}}
\][/tex]
10. Rationalize the denominator:
[tex]\[
\sin (\theta) = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}
\][/tex]
Therefore, the exact value is:
[tex]\[
\sin (\theta) = \frac{2 \sqrt{5}}{5}
\][/tex]
