To determine the sine of the angle formed by a terminal ray intersecting the unit circle at the point \((12, 3\sqrt{2})\), we first need to verify that the point lies on the unit circle.
The unit circle is defined by the equation: \[ x^2 + y^2 = 1 \]
Let's calculate \(x^2 + y^2\) for the point \((12, 3\sqrt{2})\):
\[ x^2 = 12^2 = 144 \] \[ y^2 = (3\sqrt{2})^2 = 9 \cdot 2 = 18 \]
Now, sum these values:
\[ x^2 + y^2 = 144 + 18 = 162 \]
Since \(162 \neq 1\), the point \((12, 3\sqrt{2})\) does not lie on the unit circle.
However, if you're looking for the sine of the angle formed by the terminal ray that would intersect the unit circle, we will normalize the point to get a point on the unit circle.
To normalize, we divide both coordinates by the distance from the origin to the point \((12, 3\sqrt{2})\). The distance \(r\) can be calculated as:
\[ r = \sqrt{144 + 18} = \sqrt{162} = 9\sqrt{2} \]
Now, we find the normalized coordinates:
\[ \left( \frac{12}{9\sqrt{2}}, \frac{3\sqrt{2}}{9\sqrt{2}} \right) = \left( \frac{4}{3\sqrt{2}}, \frac{1}{3} \right) \]
Now, we can find the sine of the angle, which corresponds to the \(y\)-coordinate of this normalized vector:
\[ \sin(\theta) = \frac{1}{3} \]
Thus, the sine of the angle formed by the terminal ray that intersects the unit circle at this point is:
\[ \frac{1}{3} \]
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