Problem 30 - Algebra and Trigonometry

Question

\(21-30=\) Find the terminal point \(P(x, y)\) on the unit circle determined by the given value of \(t .\) $$ t=\frac{11 \pi}{6} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Unit Circle

The unit circle is a circle with a radius of 1, centered at the origin of a coordinate plane. Every point on the circle can be described by an angle \( t \) radiating from the positive x-axis.

Step 2: Determine the Reference Angle

The given angle \( t = \frac{11\pi}{6} \) is in radians. A full circle is \( 2\pi \), so subtract \( 2\pi \) to find the reference angle within the unit circle. Thus, \( \frac{11\pi}{6} - \frac{12\pi}{6} = -\frac{\pi}{6} \). The reference angle is \( \frac{\pi}{6} \).

Step 3: Determine the Quadrant

Since \( \frac{11\pi}{6} \) is equivalent to \( 2\pi - \frac{\pi}{6} \), this angle is in the fourth quadrant of the unit circle.

Step 4: Calculate Coordinates Using Reference Angle

In the fourth quadrant, the cosine is positive, and the sine is negative. The reference angle is \( \frac{\pi}{6} \), where \( \cos{\frac{\pi}{6}} = \frac{\sqrt{3}}{2} \) and \( \sin{\frac{\pi}{6}} = \frac{1}{2} \). Therefore, \( P = \left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right) \).

Key Concepts

Reference Angle

Understanding the concept of a reference angle is key to solving trigonometric problems related to the unit circle. A reference angle is the smallest angle that a terminal side makes with the x-axis when the angle is placed in standard position.
Reference angles are always within the range of \(0\) to \(\frac{\pi}{2}\) radians (or \(0^\circ\) to \(90^\circ\)). They help simplify calculations because they allow you to use known values of trigonometric functions for specific angles like \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), and \(\frac{\pi}{3}\).
To find a reference angle, you'll subtract \(2\pi\) from the given angle to format it within the unit circle, then further adjust it to ensure it's in the first quadrant. In our exercise, the angle \(t = \frac{11\pi}{6}\) is reduced by subtracting the full circle (2\pi), resulting in a negative value \(-\frac{\pi}{6}\), indicating its placement in the fourth quadrant, but the reference angle itself is \(\frac{\pi}{6}\). This serves as a handy reference for trigonometric calculations.

Quadrants

Understanding quadrants helps in determining the behavior of angles on the unit circle. The coordinate plane is divided into four quadrants, each representing different signs for cosine and sine:

• Quadrant I: Both cosine and sine are positive.
• Quadrant II: Cosine is negative, and sine is positive.
• Quadrant III: Both cosine and sine are negative.
• Quadrant IV: Cosine is positive, and sine is negative.

In our exercise, \(t = \frac{11\pi}{6}\) is placed in the fourth quadrant. The fourth quadrant is critical because angles here reflect a positive cosine and a negative sine. Recognizing in which quadrant an angle lies is essential, as it affects the coordinate values you use. For example, though the reference angle \(\frac{\pi}{6}\) has known values of \(rac{\sqrt{3}}{2}\) for cosine and \(rac{1}{2}\) for sine, these values must be adjusted based on the quadrant, reflecting in the fourth quadrant position as \((\frac{\sqrt{3}}{2}, -\frac{1}{2})\).
Try to grasp the significance of quadrant signs, as they guide you in adjusting the signs of your trigonometric coordinates appropriately.

Trigonometric Coordinates

Trigonometric coordinates on the unit circle are derived from standard angles and their positions within different quadrants. Each point on the unit circle is characterized by a pair of coordinates \(x, y\) which represent the cosine and sine of the angle formed from the positive x-axis. Depending on the quadrant, the signs of these coordinates change.
For angles like \(\frac{11\pi}{6}\), their coordinates are determined by their reference angle and the quadrant they fall in. The reference angle \(\frac{\pi}{6}\) geometrically provides cosine and sine values of \(\frac{\sqrt{3}}{2}\) and \(\frac{1}{2}\), respectively. However, as \(t\) is in the fourth quadrant, you need to adjust for signs: cosine remains positive, the x-value, while sine becomes negative, the y-value. Thus, the coordinates become \(P = \left( \frac{\sqrt{3}}{2}, -\frac{1}{2} \right)\).
Seeing the trigonometric coordinates this way simplifies working with angles and aids in predicting directional paths and position on the circle, a crucial skill when tackling more complex trigonometric functions.