On the unit circle, find the value of $cos(135°) + sin(225°) + an(315°)$.

$ \cos(135°) $

Since $135°$ lies in the second quadrant, we have:

$ \cos(135°) = -\cos(180° – 135°) = -\cos(45°) = -\frac{\sqrt{2}}{2} $

$ \sin(225°) $

Since $225°$ lies in the third quadrant, we have:

$ \sin(225°) = -\sin(360° – 225°) = -\sin(135°) = -\sin(180° – 135°) = -\sin(45°) = -\frac{\sqrt{2}}{2} $

$ \tan(315°) $

Since $315°$ lies in the fourth quadrant, we have:

$ \tan(315°) = \tan(360° – 45°) = \tan(45°) = 1 $

Combining all these, we get:

$ \cos(135°) + \sin(225°) + \tan(315°) = -\frac{\sqrt{2}}{2} + -\frac{\sqrt{2}}{2} + 1 = -\sqrt{2} + 1 $

To find the value of $cos(135°) + sin(225°) + an(315°)$, we first calculate each trigonometric function individually:

$ cos(135°) = -frac{sqrt{2}}{2}, sin(225°) = -frac{sqrt{2}}{2}, an(315°) = 1 $

Next, we sum the values:

$ cos(135°) + sin(225°) + an(315°) = -frac{sqrt{2}}{2} + -frac{sqrt{2}}{2} + 1 $

Simplifying:

$ -frac{sqrt{2}}{2} + -frac{sqrt{2}}{2} = -sqrt{2} $

Hence, the final result is:

$ -sqrt{2} + 1 $

The value can be calculated as follows:

$ cos(135°) = -frac{sqrt{2}}{2}, sin(225°) = -frac{sqrt{2}}{2}, an(315°) = 1 $

Summing the values:

$ -frac{sqrt{2}}{2} – frac{sqrt{2}}{2} + 1 = -sqrt{2} + 1 $

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