Certainly! Let's solve for [tex]\(\sin(\theta)\)[/tex] given that [tex]\(\cos(\theta) = -\frac{2\sqrt{5}}{5}\)[/tex] and [tex]\(\theta\)[/tex] is in quadrant II.
Since [tex]\(\theta\)[/tex] is in quadrant II, we know that the cosine value is negative and the sine value is positive.
We can use the Pythagorean identity:
[tex]\[
\sin^2(\theta) + \cos^2(\theta) = 1
\][/tex]
First, plug in the given value of [tex]\(\cos(\theta)\)[/tex]:
[tex]\[
\cos^2(\theta) = \left(-\frac{2\sqrt{5}}{5}\right)^2
\][/tex]
Calculate [tex]\(\cos^2(\theta)\)[/tex]:
[tex]\[
\cos^2(\theta) = \left(-\frac{2\sqrt{5}}{5}\right)^2 = \left(\frac{2\sqrt{5}}{5}\right)^2 = \frac{4 \times 5}{25} = \frac{20}{25} = \frac{4}{5}
\][/tex]
Now, use the Pythagorean identity to find [tex]\(\sin^2(\theta)\)[/tex]:
[tex]\[
\sin^2(\theta) = 1 - \cos^2(\theta)
\][/tex]
[tex]\[
\sin^2(\theta) = 1 - \frac{4}{5} = \frac{5}{5} - \frac{4}{5} = \frac{1}{5}
\][/tex]
We need [tex]\(\sin(\theta)\)[/tex], so take the square root of both sides:
[tex]\[
\sin(\theta) = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}
\][/tex]
Since [tex]\(\theta\)[/tex] is in quadrant II where the sine value is positive, we get:
[tex]\[
\sin(\theta) = \frac{\sqrt{5}}{5}
\][/tex]
Thus, the correct answer is:
B. [tex]\(\frac{\sqrt{5}}{5}\)[/tex]
