Problem 23 - Algebra and Trigonometry

Question

\(21-30=\) Find the terminal point \(P(x, y)\) on the unit circle determined by the given value of \(t .\) $$ t=\frac{5 \pi}{6} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Unit Circle

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

Step 2: Determine Reference Angle

Firstly, find the reference angle for \( t = \frac{5\pi}{6} \). Since the angle is in the second quadrant, its reference angle can be calculated as \( \pi - \frac{5\pi}{6} = \frac{\pi}{6} \).

Step 3: Find Coordinates Using Sine and Cosine Values

The coordinates for \( t = \frac{5\pi}{6} \) can be determined using the reference angle \( \frac{\pi}{6} \) and the fact that cosine and sine values have specific signs in the second quadrant. \( \cos\left(\frac{5\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2} \) and \( \sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \).

Step 4: Write the Terminal Point

The terminal point \( P(x, y) \) on the unit circle for \( t = \frac{5\pi}{6} \) is \( P\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right) \).

Key Concepts

Reference Angle

A reference angle is a useful concept when working with angles on the unit circle. It allows us to determine an equivalent angle that is always between 0 and \(\frac{\pi}{2}\) (90°). This angle helps to simplify calculations by using the fundamental properties of the angles in the first quadrant. To find the reference angle for an angle \(t\), we look at where \(t\) lies in the coordinate plane. For example:

• If \(t\) is in the second quadrant, we calculate the reference angle as \(\pi - t\).
• If \(t\) is in the third quadrant, it is \(t - \pi\).
• In the fourth quadrant, it's computed as \(2\pi - t\).

In the context of our problem, the angle \(t = \frac{5\pi}{6}\) is in the second quadrant. Thus, its reference angle is \(\pi - \frac{5\pi}{6} = \frac{\pi}{6}\). This reference angle corresponds to an angle in the first quadrant, enabling us to conveniently use known sine and cosine values for calculations.

Coordinates of Points

The coordinates of any point on the unit circle are crucial because they represent the sine and cosine of the angle \(t\) from the positive x-axis. Every angle \(t\) on the circle corresponds to a unique point \(P(x, y)\) where \(x = \cos(t)\) and \(y = \sin(t)\). Here’s how we determine which trig values to use based on the angle’s location on the unit circle:

• In the first quadrant, both \(x\) and \(y\) are positive.
• In the second quadrant, \(x\) is negative and \(y\) is positive.
• In the third quadrant, both \(x\) and \(y\) are negative.
• In the fourth quadrant, \(x\) is positive and \(y\) is negative.

For the angle \(t = \frac{5\pi}{6}\), which is in the second quadrant, the coordinates are determined by considering the negative cosine and positive sine of its reference angle \(\frac{\pi}{6}\). Therefore, \(P(x, y) = P\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\).

Sine and Cosine Values

Sine and cosine play a vital role in determining the exact coordinates of points on the unit circle. Both values are dependent on the reference angle and the quadrant in which the original angle \(t\) is located. In our specific case, the reference angle for \(t = \frac{5\pi}{6}\) is \(\frac{\pi}{6}\). The primary sine and cosine values for \(\frac{\pi}{6}\) in the first quadrant are:

• \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\)
• \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)

Since the original angle resides in the second quadrant, the signs of the trigonometric ratios are adjusted accordingly: cosine is negative, while sine remains positive. So,

• \(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\)
• \(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\)

These values are key to finding the terminal point \(P(x, y) = P\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\) on the unit circle.