Problem 24 - Algebra and Trigonometry

Question

\(21-30=\) Find the terminal point \(P(x, y)\) on the unit circle determined by the given value of \(t .\) $$ t=\frac{7 \pi}{6} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Unit Circle

The unit circle is a circle with a radius of 1 unit, centered at the origin (0,0) in the coordinate plane. A terminal point on the unit circle for an angle \( t \) from the positive x-axis is given by \((\cos t, \sin t)\).

Step 2: Simplify the Angle

The given angle is \( t = \frac{7 \pi}{6} \). Since this is expressed in radians, we first confirm it lies between \( 0 \) and \( 2\pi \). This angle is in the third quadrant because \( \pi < \frac{7\pi}{6} < \frac{3\pi}{2} \).

Step 3: Calculate Cosine and Sine

In the unit circle, for any angle \( t \), we use its reference angle to find coordinates. The reference angle \( \theta \) for \( \frac{7\pi}{6} \) is \( \frac{7\pi}{6} - \pi = \frac{\pi}{6} \). For \( \theta = \frac{\pi}{6} \), \( \cos \theta = \frac{\sqrt{3}}{2} \) and \( \sin \theta = \frac{1}{2} \). In the third quadrant, cosine and sine are negative; thus \( \cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2} \) and \( \sin \frac{7\pi}{6} = -\frac{1}{2} \).

Step 4: Write the Terminal Point

The terminal point \( P(x, y) \) determined by \( t = \frac{7\pi}{6} \) on the unit circle is \( P\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right) \).

Key Concepts

Terminal Point

The terminal point on the unit circle is the final position of a point that travels counterclockwise around the circle starting from the positive x-axis. It gives us a direct way to identify the coordinates of a point on the circle for a specific angle \( t \). The coordinates of the terminal point are found using trigonometric functions, namely cosine and sine.
- If \( t \) is expressed in radians, as it often is, you'll use \((\cos t, \sin t)\) to determine the terminal point.This provides a standardized way of reflecting angles and is especially helpful for solving trigonometry problems. In the exercise above, the terminal point for \( t = \frac{7\pi}{6} \) is given by the coordinates \( (-\frac{\sqrt{3}}{2}, -\frac{1}{2}) \), which reflects the unit circle's position in the third quadrant.

Radian Measure

Radian measure is a way of expressing angles using the radius of a circle. It's a natural way to think about angles because it relates directly to the geometry of the circle itself. One full revolution around a circle is \(2\pi\) radians, which is equivalent to 360 degrees.
- To convert degrees to radians, you multiply by \(\frac{\pi}{180}\).- To convert radians to degrees, you multiply by \(\frac{180}{\pi}\).
For the given angle \( t = \frac{7\pi}{6} \), it is already in radians, so there is no need for conversion. Radian measure allows for more straightforward calculations since it aligns calculations with the unit circle's inherent proportions.

Trigonometric Functions

The trigonometric functions \( \cos \) and \( \sin \) are pivotal for locating points on the unit circle. These functions correspond to the x and y coordinates of a point on the unit circle. Using the coordinates \((\cos t, \sin t)\), these functions allow us to evaluate angles precisely and are defined as follows:

• Cosine \(\cos t\) measures the horizontal distance from the origin to the point on the unit circle.
• Sine \(\sin t\) measures the vertical distance from the origin to that same point.

In the example of \( t = \frac{7\pi}{6} \), both \( \cos \left( \frac{7\pi}{6} \right)\) and \( \sin \left( \frac{7\pi}{6} \right) \) are negative, reflecting the location in the third quadrant, where both x and y values are negative.

Reference Angle

A reference angle is the smallest angle that the given angle makes with the x-axis, providing a simpler angle with which to work. It helps in deducing the sine and cosine values for angles greater than 90 degrees or \(\pi/2\) radians. To compute a reference angle for an angle \(t\):
- If \(t\) is greater than \(\pi\), subtract \(\pi\) to obtain the reference angle.
For \(t = \frac{7\pi}{6}\), its reference angle is \(\frac{7\pi}{6} - \pi = \frac{\pi}{6}\).While the reference angle preserves the magnitude of the trigonometric functions, the signs of \(\cos t\) and \(\sin t\) depend on the quadrant in which \(t\) resides. Thus, we use reference angles to ensure correct calculations involving angles greater than \(\pi/2\).