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The trigonometric function "sine," often abbreviated as \( \sin \), is one of the three primary functions in trigonometry. When given a point \( P(x, y) \), the sine of an angle \( t \), which is determined by the point, is used to describe the vertical position (or height) relative to the unit circle.
To find \( \sin t \), we use the formula:

• \( \sin t = \frac{y}{r} \)

In this context, \( r \) represents the radius of the circle from the origin to the given point, essentially the hypotenuse if you consider a right-angled triangle formed along the axes.
For the given point \( \left(-\frac{1}{3}, \frac{2\sqrt{2}}{3}\right) \), we determined that \( r = 1 \), hence \( \sin t = \frac{2\sqrt{2}}{3} \). This calculation indicates that for the terminal point on the circle, the vertical position is \( \frac{2\sqrt{2}}{3} \), reflecting the height proportional to its radius.

The cosine function, written as \( \cos \), is another of the core trigonometric functions. It reflects the horizontal distance from the origin to a point \( P(x, y) \) on the unit circle. The formula for determining \( \cos t \) is similar in structure to sine:

• \( \cos t = \frac{x}{r} \)

Here, \( x \) is the x-coordinate of the point, and \( r \) is again the radius of the unit circle. The unit circle concept assumes that \( r = 1 \), simplifying many calculations.
For \( P\left(-\frac{1}{3}, \frac{2\sqrt{2}}{3}\right) \), substituting in the value for \( x \) and \( r = 1 \) gives us \( \cos t = -\frac{1}{3} \). This value tells you how far the point is horizontally on the unit circle, and in this case, it lies to the left of the origin.

The tangent of an angle, expressed as \( \tan \), is a function that relates the sine and cosine values. It effectively describes the slope or the angle of a line extending from the origin through the point \( P(x, y) \). The formula to find \( \tan t \) is:

• \( \tan t = \frac{y}{x} \)

This equation shows a ratio comparing the vertical coordinate to the horizontal coordinate of the point. In simpler terms, it quantifies how steeply a line runs from the origin through the point.
For this problem, using the coordinates \( \left(-\frac{1}{3}, \frac{2\sqrt{2}}{3}\right) \) results in a tangent value of \( \tan t = -2\sqrt{2} \). This value is negative, indicating that the line descends from left to right, because the x-coordinate is negative while the y-coordinate is positive.