Find the coordinates of the point where the terminal side of a $225^{circ}$ angle intersects the unit circle.

Given an angle of $225^{\circ}$, we first convert it to radians:

$225^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{5\pi}{4} \text{ radians}$

The angle $\frac{5\pi}{4}$ is in the third quadrant, where both sine and cosine are negative. The reference angle is $225^{\circ} – 180^{\circ} = 45^{\circ}$. Since $\sin(45^{\circ}) = \frac{\sqrt{2}}{2}$ and $\cos(45^{\circ}) = \frac{\sqrt{2}}{2}$, we have:

$\sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$

$\cos(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$

Thus, the coordinates are:

$(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2})$

To find the intersection point of the terminal side of a $225^{circ}$ angle with the unit circle, we convert the angle to radians:

$225^{circ} = frac{5pi}{4} ext{ radians}$

Next, we note that $frac{5pi}{4}$ is in the third quadrant. For a reference angle of $45^{circ}$, we know:

$cos(45^{circ}) = frac{sqrt{2}}{2}$

$sin(45^{circ}) = frac{sqrt{2}}{2}$

Since both cosine and sine are negative in the third quadrant:

$cos(frac{5pi}{4}) = -frac{sqrt{2}}{2}$

$sin(frac{5pi}{4}) = -frac{sqrt{2}}{2}$

The coordinates are:

$(-frac{sqrt{2}}{2}, -frac{sqrt{2}}{2})$

Convert $225^{circ}$ to radians:

$225^{circ} = frac{5pi}{4}$

In the third quadrant:

$cos(frac{5pi}{4}) = -frac{sqrt{2}}{2}$

$sin(frac{5pi}{4}) = -frac{sqrt{2}}{2}$

Coordinates:

$(-frac{sqrt{2}}{2}, -frac{sqrt{2}}{2})$

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