Problem 36 - Precalculus

Question

Use the unit circle to find all of the exact values of \(\theta\) that make the equation true in the indicated interval. $$\cos \theta=-\frac{\sqrt{3}}{2}, 0 \leq \theta \leq 2 \pi$$

Step-by-Step Solution

Answer

The solutions are \( \theta = \frac{5\pi}{6} \) and \( \theta = \frac{7\pi}{6} \).

Step 1: Understanding the Problem

We need to find all angles \( \theta \) within the interval \( [0, 2\pi] \) for which the cosine value equals \(-\frac{\sqrt{3}}{2}\). This means finding the angles on the unit circle where the x-coordinate (cosine) is \(-\frac{\sqrt{3}}{2}\).

Step 2: Identifying the Reference Angle

The reference angle whose cosine is \(\frac{\sqrt{3}}{2}\) is \(\frac{\pi}{6}\) (or 30 degrees), since cosine of \(30^\circ\) is \(\frac{\sqrt{3}}{2}\). We look for angles in the unit circle whose cosine is the negative of this value.

Step 3: Finding Corresponding Angles in Quadrants II and III

Since cosine is negative in the second and third quadrants, we find \(\theta\) in these quadrants. In Quadrant II, the angle is \(\pi - \frac{\pi}{6} = \frac{5\pi}{6} \). In Quadrant III, the angle is \(\pi + \frac{\pi}{6} = \frac{7\pi}{6} \).

Step 4: Listing All Solutions in the Interval

The angles that satisfy \( \cos \theta = -\frac{\sqrt{3}}{2} \) within \([0, 2\pi]\) are \( \theta = \frac{5\pi}{6} \) and \( \theta = \frac{7\pi}{6} \).

Key Concepts

Cosine Function

The cosine function is a fundamental part of trigonometry, often related to the measurement of angles. In a unit circle, cosine represents the x-coordinate of a point at a given angle from the origin. When we explore the unit circle, the radius is always 1, which simplifies finding cosine values as they are directly the x-coordinate.
- **Cos(x)**: Measures the horizontal distance from the center of the circle to the edge point where the angle \( \theta \) ends. - **Understanding Cosine in Trigonometric Functions**: It helps in determining the relationship between angles and sides in right triangles.
For example, the task at hand asks us to find when cosine equals \( -\frac{\sqrt{3}}{2} \), which points us to specific positions on the unit circle where these x-coordinates exist. Knowing that cosine is negative in certain quadrants helps us find the solutions.

Reference Angle

A reference angle is the smallest angle that the terminal side of any angle makes with the x-axis. It is always measured in the positive direction and is always between \( 0 \) and \( \frac{\pi}{2} \) radians (or between 0 and 90 degrees).
To solve the problem, it is crucial to recognize that the reference angle helps identify angles with equivalent trigonometric ratios in other quadrants. In this case, the cosine of the reference angle \( \frac{\pi}{6} \) is \( \frac{\sqrt{3}}{2} \). However, since cosine is negative, our task is to find where it becomes \( -\frac{\sqrt{3}}{2} \).

• The reference angle determines which angles in other quadrants will have the same absolute value of cosine.
• Use it to compute the actual angles in Quadrants II and III by adjusting the initial position accordingly.

Thus, recognizing the reference angle allows us to measure equivalent but appropriately signed cosine values at \( \frac{5\pi}{6} \) and \( \frac{7\pi}{6} \).

Quadrants

Understanding quadrants in the context of the unit circle aids in finding the sign of trigonometric functions at given angles. The unit circle is divided into four quadrants:

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

To solve the problem of finding angles with a specific cosine value of \( -\frac{\sqrt{3}}{2} \), you look at Quadrants II and III where cosine is negative:- **In Quadrant II**, angles are found by subtracting the reference angle from \( \pi \). Hence, \( \pi - \frac{\pi}{6} = \frac{5\pi}{6} \).- **In Quadrant III**, you add the reference angle to \( \pi \), giving \( \pi + \frac{\pi}{6} = \frac{7\pi}{6} \).
Understanding how the signs of trigonometric functions vary across these quadrants helps us to solve equations involving trigonometric identities efficiently.