To find the cosine and tangent of the angle given that \(\sin(\theta) = \frac{\sqrt{5}}{5}\) and that the angle \(\theta\) is in the second quadrant, we can use the following identities:
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The Pythagorean identity: \[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
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The definition of tangent: \[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]
Step 1: Find Cosine
Using the Pythagorean identity:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Plugging in the known value of sine:
\[ \left(\frac{\sqrt{5}}{5}\right)^2 + \cos^2(\theta) = 1 \] \[ \frac{5}{25} + \cos^2(\theta) = 1 \] \[ \frac{1}{5} + \cos^2(\theta) = 1 \]
Subtracting \(\frac{1}{5}\) from both sides:
\[ \cos^2(\theta) = 1 - \frac{1}{5} \] \[ \cos^2(\theta) = \frac{5}{5} - \frac{1}{5} = \frac{4}{5} \]
Now take the square root to find cosine:
\[ \cos(\theta) = \pm \sqrt{\frac{4}{5}} = \pm \frac{2}{\sqrt{5}} = \pm \frac{2\sqrt{5}}{5} \]
Since \(\theta\) is in the second quadrant, where cosine is negative, we have:
\[ \cos(\theta) = -\frac{2\sqrt{5}}{5} \]
Step 2: Find Tangent
Now we can find the tangent using the sine and cosine we found:
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]
Plugging in the values:
\[ \tan(\theta) = \frac{\frac{\sqrt{5}}{5}}{-\frac{2\sqrt{5}}{5}} \]
The \(\frac{5}{5}\) cancels out:
\[ \tan(\theta) = \frac{\sqrt{5}}{-2\sqrt{5}} = -\frac{1}{2} \]
Summary
So the values are:
- \(\cos(\theta) = -\frac{2\sqrt{5}}{5}\)
- \(\tan(\theta) = -\frac{1}{2}\)