Problem 40 - Precalculus

Question

Use the unit circle to find all of the exact values of \(\theta\) that make the equation true in the indicated interval. $$\sin \theta=-1,0 \leq \theta \leq 4 \pi$$

Step-by-Step Solution

Answer

\(\theta = \frac{3\pi}{2}\) and \(\theta = \frac{7\pi}{2}\) are the solutions.

Step 1: Understand the Problem

We need to find all angle values \(\theta\) where the sine of the angle equals -1, given that \(\theta\) is within the range from 0 to \(4\pi\).

Step 2: Use the Unit Circle Concept

Recall that the sine of an angle corresponds to the \(y\)-coordinate of the point where the angle intersects the unit circle, which has a radius of 1.

Step 3: Identify Where \(\sin \theta = -1\) on the Unit Circle

On the unit circle, the \(\sin \theta = -1\) occurs at the angle \(\theta = \frac{3\pi}{2}\) since the \(y\)-coordinate at this position is -1. The circle completes a full loop at \(2\pi\).

Step 4: Consider the Interval \([0, 4\pi]\)

Since \(0 \leq \theta \leq 4\pi\), check for additional occurrences of \(\theta = \frac{3\pi}{2}\) plus any full rotations. A full rotation is \(2\pi\), so subsequent solutions are \(\frac{3\pi}{2} + 2\pi k\) where \(k\) is an integer.

Step 5: Calculate Additional Values within the Given Interval

- For \(k = 0\), \(\theta = \frac{3\pi}{2}\). - For \(k = 1\), \(\theta = \frac{3\pi}{2} + 2\pi \times 1 = \frac{7\pi}{2}\). - For \(k = 2\), \(\theta = \frac{3\pi}{2} + 2\pi \times 2 = \frac{11\pi}{2}\). Since \(\frac{11\pi}{2} > 4\pi\), check with a decimal approximation that \(11\pi/2 \approx 17.28/2\), this is indeed greater than \(4\pi\approx 12.56\). Therefore, this is outside our interval of interest.

Step 6: List the Valid Solutions

The values within the specified interval are \(\theta = \frac{3\pi}{2}\) and \(\theta = \frac{7\pi}{2}\).

Key Concepts

Sine Function

The sine function is a fundamental concept in trigonometry, represented as \( \sin(\theta) \) where \( \theta \) is the angle. It relates to a right triangle, representing the ratio of the length of the opposite side to the hypotenuse. On the unit circle, it corresponds to the \( y \)-coordinate of a point where the angle's terminal side intersects the circle.

Key characteristics of the sine function include:

• Periodic nature with a period of \( 2\pi \)
• Range of values between -1 and 1
• Symmetry around the origin, which makes \( \sin(-\theta) = -\sin(\theta) \)

The sine function is used extensively in solving various trigonometric equations, particularly when trying to find angles with specific sine values.

Angle Measurement

Angles are measured in degrees or radians, but radians are more commonly used in mathematics because they provide a direct relationship with the properties of circles. One full rotation around a circle equals \( 2\pi \) radians, equivalent to 360 degrees.

The unit circle is a circle with radius 1, centered at the origin of a coordinate system. It offers a convenient way to measure angles because:

• Angles start from the positive \( x \)-axis and rotate counterclockwise.
• The circle's circumference is \( 2\pi \), allowing us to measure angles in radians.
• The coordinates of points on the unit circle directly relate to trigonometric functions.

In the context of the exercise, measuring angles concerning the unit circle helps us find where specific sine values, such as -1, are located.

Trigonometric Equations

Trigonometric equations involve trigonometric functions and are equations like \( \sin(\theta) = -1 \) that we solve for \( \theta \). Solving these equations often involves:

• Using properties of the trigonometric functions
• Employing the unit circle to find angles
• Considering the periodic nature to find all possible solutions

In your exercise, \( \sin(\theta) = -1 \) represents a specific \( y \)-coordinate on the unit circle where the angle \( \theta \) intersects the circle. Since the sine function is periodic, we consider multiple solutions by rotating full circles, or \( 2\pi \), from the primary solution \( \theta = \frac{3\pi}{2} \). This process ensures all solutions within the interval \([0, 4\pi]\) are found.