Question

macy is told that an angle has a sine of √5/5 and is in quadrant 2 find cosine and tangent using identities and the quadrant

Answer 1

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:

1. The Pythagorean identity:
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]

2. 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:

- \(\cos(\theta) = -\frac{2\sqrt{5}}{5}\)
- \(\tan(\theta) = -\frac{1}{2}\)