Problem 27 - Precalculus Mathematics for Calculus

Question

Find the terminal point \(P(x, y)\) on the unit circle determined by the given value of \(t .\) $$t=-\frac{\pi}{3}$$

Step-by-Step Solution

Answer

The terminal point is \(\left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right)\).

Step 1: Understanding the Problem

We need to find the point \(P(x, y)\) on the unit circle that corresponds to the angle \(t = -\frac{\pi}{3}\). The unit circle is a circle with radius 1 centered at the origin.

Step 2: Determining the Angle on the Unit Circle

Since \(t = -\frac{\pi}{3}\), we are dealing with a negative angle. Negative angles are measured clockwise from the positive x-axis. In other words, we need to move \(\frac{\pi}{3}\) radians clockwise from the positive x-axis.

Step 3: Finding the Reference Angle

For the angle \(-\frac{\pi}{3}\), the reference angle, which is the acute angle the terminal side makes with the x-axis, is \(\frac{\pi}{3}\). This will help us determine the x and y coordinates on the unit circle.

Step 4: Calculating the Coordinates

The cosine of the reference angle \(\frac{\pi}{3}\) is \(\frac{1}{2}\) and the sine is \(\frac{\sqrt{3}}{2}\). Since the angle is moving clockwise, it places it in the fourth quadrant, where cosine is positive and sine is negative. Thus, \(x = \cos(-\frac{\pi}{3}) = \frac{1}{2}\) and \(y = \sin(-\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}\).

Step 5: Writing the Terminal Point

The terminal point \(P(x, y)\) on the unit circle for \(t = -\frac{\pi}{3}\) is therefore \(\left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right)\).

Key Concepts

Terminal Point

A terminal point on the unit circle is the precise location where the angle ends as it rotates from the initial side along the circle. The initial side is typically along the positive x-axis. Understanding the terminal point is fundamental in trigonometry because it helps us determine the specific coordinates

• belonging to a given angle.

In our exercise, we are tasked to find the terminal point for the angle \(t = -\frac{\pi}{3} \). Since it is a negative angle, the rotation direction will be clockwise. This leads us to land in the fourth quadrant of the unit circle, where the terminal point ultimately helps in finding out specific coordinates of a trigonometric angle like cosine and sine. In trigonometric functions, knowing the terminal point confirms the x and y values at which the angle settles on the circle, helping us in calculations and understanding more complex trig functions.

Reference Angle

A reference angle is an acute angle that a given angle makes with the x-axis. It is always positive and is found within a triangle formed by dropping a perpendicular to the x-axis. The reference angle provides an easy way to handle angles across different quadrants by referring back to the first quadrant values.
In the case of our angle, \(-\frac{\pi}{3}\), its reference angle is \(\frac{\pi}{3}\).This is because we are examining the absolute acute angle from the axis without considering direction. Reference angles are crucial because they allow us to apply known trigonometric values,

• such as sine or cosine, to any angle,
• even those outside of the first quadrant.

This is especially useful in unit circle calculations since trigonometric values found using reference angles translate directly into finding specific coordinate values for angles.

Coordinates on the Unit Circle

Coordinates on the unit circle are the \((x, y)\) pairs representing points formed by angles as they rotate from the x-axis. In trigonometry, these coordinates correlate directly with the cosine and sine values of the angle.
For a unit circle with its center at \((0, 0)\) and radius 1, the x-coordinate is found using \(\cos(t)\) and the y-coordinate with \(\sin(t)\).Given the reference angle \(\frac{\pi}{3}\)in the solution, we use this information to find:

• Cosine, which gives the x-coordinate as \(x = \cos(-\frac{\pi}{3}) = \frac{1}{2}\).
• Sine, yielding the y-coordinate as \(y = \sin(-\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}\).

Since \(-\frac{\pi}{3}\) places us in the fourth quadrant, cosine remains positive, while sine switches to negative, resulting in the coordinates \(\left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right)\). Understanding these coordinate positions on the unit circle is key in graphing and solving trigonometric functions effectively.