$ ext{Find the value of } heta ext{ if } an( heta) = 2 ext{ and } heta ext{ is in the second quadrant. Then, calculate the coordinates of the corresponding point on the unit circle.} $

We know that \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). In the second quadrant, sine is positive and cosine is negative. Let us find \( \theta \) such that \( \tan(\theta) = 2 \). The reference angle \( \theta_r \) is given by:

$ \theta_r = \arctan(2) $

Since \( \theta \) is in the second quadrant, the angle \( \theta \) is:

$ \theta = \pi – \theta_r = \pi – \arctan(2) $

Next, to find the coordinates of the corresponding point on the unit circle, we use the unit circle property \((\cos(\theta), \sin(\theta))\). First we find \( \sin(\theta) \) and \( \cos(\theta) \) using:

$ \sin(\theta) = \frac{2}{\sqrt{1 + 2^2}} = \frac{2}{\sqrt{5}} \quad \text{and} \quad \cos(\theta) = -\frac{1}{\sqrt{1 + 2^2}} = -\frac{1}{\sqrt{5}} $

Therefore, the coordinates are:

$ \left( -\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) $

The given problem states that ( an( heta) = 2 ) and ( heta ) is in the second quadrant. We start by recognizing that ( an heta = frac{sin heta}{cos heta} ).

To solve for ( heta ), we first find the reference angle ( heta_r ):

$ heta_r = arctan(2) $

In the second quadrant, we modify ( heta_r ) to find the actual angle ( heta ):

$ heta = pi – heta_r = pi – arctan(2) $

Using ( sin( heta) ) and ( cos( heta) ) properties, we have:

$ sin( heta) = 2 / sqrt{5} quad ext{and} quad cos( heta) = -1 / sqrt{5} $

Thus, on the unit circle, the coordinates are:

$ left( -1/sqrt{5}, 2/sqrt{5}
ight) $

Given ( heta ) in the second quadrant with ( an( heta) = 2 ), we find ( heta ) as:

$ heta = pi – arctan(2) $

Coordinates on the unit circle are:

$ left( -frac{1}{sqrt{5}}, frac{2}{sqrt{5}}
ight) $

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