Problem 37 - Precalculus

Question

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

Step-by-Step Solution

Answer

\(\theta = \frac{4\pi}{3}\) and \(\theta = \frac{5\pi}{3}\).

Step 1: Identify Quadrants for \\(\sin \\\theta < 0\\)

Since \(\sin \, \theta = -\frac{\sqrt{3}}{2}\), we are looking for angles where sine is negative. In the unit circle, sine is negative in the third and fourth quadrants.

Step 2: Determine Reference Angle for Sine

The reference angle is the angle which satisfies \(\sin \, \text{ref angle} = \frac{\sqrt{3}}{2}\). From knowledge of special angles, \(\text{ref angle} = \frac{\pi}{3}\) because \(\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\).

Step 3: Find Angles in the Third Quadrant

In the third quadrant, the angle \(\theta\) is given by \(\pi + \text{ref angle}\). Therefore, \(\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}\).

Step 4: Find Angles in the Fourth Quadrant

In the fourth quadrant, \(\theta\) is given by \(2\pi - \text{ref angle}\). Therefore, \(\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\).

Step 5: Verify Solutions Are Within the Interval

Both angles found, \(\frac{4\pi}{3}\) and \(\frac{5\pi}{3}\), are within the given interval \([0, 2\pi]\). Therefore, they are the correct solutions.

Key Concepts

Special Angles

Special angles are particular angles on the unit circle that have well-known sine, cosine, and tangent values. They include angles like \(0\), \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\), \(\frac{\pi}{2}\), and their equivalents in other quadrants. These angles are essential in solving trigonometric problems because they have easily recognizable values that make calculations straightforward. For instance, let's consider \(\sin(\frac{\pi}{3})\). The value is \(\frac{\sqrt{3}}{2}\). This means whenever you encounter \(\sin \theta = \frac{\sqrt{3}}{2}\), you should immediately think of \(\frac{\pi}{3}\) or its related angles in other quadrants. This knowledge helps in quickly identifying reference angles and solving trigonometric equations. Knowing these special angles allows for ease of calculation, especially when handling equations that arise from the unit circle.

Trigonometric Functions

Trigonometric functions include sine, cosine, and tangent, each with its unique characteristics on the unit circle. These functions relate angles to the ratios of the sides of a right triangle.

• Sine: This function provides the y-coordinate of the point on the unit circle. It has a value ranging from -1 to 1.
• Cosine: Similar to sine but provides the x-coordinate, also ranging from -1 to 1.
• Tangent: Represents the ratio of the sine to the cosine, somewhat different because it can take on any real number value.

Understanding where these functions are positive or negative is key. For instance, in the unit circle, the sine function is negative in the third and fourth quadrants, meaning that any angle beyond \(\pi\) (180 degrees) but less than \(2\pi\) (360 degrees) will yield a negative sine value. This is crucial when solving trigonometric equations within a specified interval, as in the original exercise.

Reference Angle

The reference angle is the smallest angle between the terminal side of your given angle and the x-axis in the unit circle. It is always a positive acute angle—less than \(90^\circ\) or \(\pi/2\). Reference angles are valuable because they allow us to compute the trigonometric functions for any angle, using only the values from the first quadrant.

Calculating Reference Angles:

For instance, for the exercise \(\sin \theta = -\frac{\sqrt{3}}{2}\), first focus on \(\sin\) positive value \(\frac{\sqrt{3}}{2}\). The angle \(\theta\) that results in this is \(\frac{\pi}{3}\), making this fraction our reference angle.From this reference angle, you can find equivalent angles in different quadrants:

• In the third quadrant: \(\theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}\).
• In the fourth quadrant: \(\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\).

These calculations help you determine the points on the unit circle and facilitate finding all potential solutions within a given interval.