To solve for [tex]\(\sin(\theta)\)[/tex] when [tex]\(\cos(\theta) = -\frac{2\sqrt{5}}{5}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant II, follow these steps:
1. Identify [tex]\(\cos(\theta)\)[/tex]: We are given that [tex]\(\cos(\theta) = -\frac{2\sqrt{5}}{5}\)[/tex]. This is a negative value, which is appropriate since cosine is negative in quadrant II.
2. Use the Pythagorean identity: The Pythagorean identity states that for any angle [tex]\(\theta\)[/tex],
[tex]\[
\sin^2(\theta) + \cos^2(\theta) = 1
\][/tex]
3. Calculate [tex]\(\cos^2(\theta)\)[/tex]:
[tex]\[
\cos^2(\theta) = \left(-\frac{2\sqrt{5}}{5}\right)^2 = \left(\frac{2\sqrt{5}}{5}\right)^2 = \frac{4 \cdot 5}{25} = \frac{20}{25} = \frac{4}{5}
\][/tex]
4. Find [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[
\sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{1}{5}
\][/tex]
5. Determine [tex]\(\sin(\theta)\)[/tex]:
[tex]\[
\sin(\theta) = \sqrt{\sin^2(\theta)} = \sqrt{\frac{1}{5}} = \frac{\sqrt{1}}{\sqrt{5}} = \frac{1}{\sqrt{5}}
\][/tex]
6. Rationalize the denominator:
[tex]\[
\sin(\theta) = \frac{1}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{5}}{5}
\][/tex]
7. Sign of [tex]\(\sin(\theta)\)[/tex] in quadrant II: In quadrant II, sine is positive. Therefore,
[tex]\[
\sin(\theta) = \frac{\sqrt{5}}{5}
\][/tex]
So, the correct value for [tex]\(\sin(\theta)\)[/tex] in this context is [tex]\(\frac{\sqrt{5}}{5}\)[/tex].
The correct answer is:
C. [tex]\(\frac{\sqrt{5}}{5}\)[/tex]
