Given that $ an( heta) = 2 $ and $ heta $ is in the second quadrant, find the exact values of $ sin( heta) $ and $ cos( heta) $.

1. Given that $ \tan(\theta) = 2 $, we can write:

$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = 2 $

Let $ \sin(\theta) = 2k $ and $ \cos(\theta) = -k $ (since $ \theta $ is in the second quadrant where cosine is negative). Then:

$ \frac{2k}{-k} = 2 $

2. From the Pythagorean identity:

$ \sin^2(\theta) + \cos^2(\theta) = 1 $

Substitute the values:

$ (2k)^2 + (-k)^2 = 1 $

$ 4k^2 + k^2 = 1 $

3. Solving for $ k $:

$ 5k^2 = 1 $

$ k^2 = \frac{1}{5} $

$ k = \pm \frac{1}{\sqrt{5}} $

4. Since $ \sin(\theta) = 2k $ and $ \cos(\theta) = -k $, we have:

$ \sin(\theta) = 2 \left( \frac{1}{\sqrt{5}} \right) = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} $

$ \cos(\theta) = – \left( \frac{1}{\sqrt{5}} \right) = -\frac{\sqrt{5}}{5} $

1. Given $ an( heta) = 2 $ and $ heta $ is in the second quadrant:

$ an( heta) = frac{sin( heta)}{cos( heta)} = 2 $

Let $ sin( heta) = 2k $ and $ cos( heta) = -k $.

2. Using the identity:

$ (2k)^2 + (-k)^2 = 1 $

$ 4k^2 + k^2 = 1 $

$ 5k^2 = 1
ightarrow k^2 = frac{1}{5} $

3. Thus:

$ sin( heta) = frac{2}{sqrt{5}} = frac{2sqrt{5}}{5} $

$ cos( heta) = -frac{1}{sqrt{5}} = -frac{sqrt{5}}{5} $

1. Given $ an( heta) = 2 $:

$ an( heta) = frac{sin( heta)}{cos( heta)} $

2. Let $ sin( heta) = 2k $ and $ cos( heta) = -k $:

$ 4k^2 + k^2 = 1
ightarrow k^2 = frac{1}{5} $

3. Therefore:

$ sin( heta) = frac{2sqrt{5}}{5} $

$ cos( heta) = -frac{sqrt{5}}{5} $

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