Problem 27 - Algebra 2

Question

Solve each equation for \(0 \leq \theta<2 \pi\). $$ (\sin \theta-1)(\sin \theta+1)=0 $$

Step-by-Step Solution

Answer

The solutions to the equation are \( \theta = \pi/2 \) and \( \theta = 3\pi/2 \).

Step 1: Set each factor equal to zero

First, we must take each factor in the original equation and set it equal to zero. This results in two equations: \n1. \( \sin(\theta) - 1 = 0 \) \n2. \( \sin(\theta) + 1 = 0 \)

Step 2: Solve each equation

Now, solve each of the equations obtained in step 1 for \( \theta \). \n1. For \( \sin(\theta) - 1 = 0 \), adding 1 to both sides gives \( \sin(\theta) = 1 \). Using the inverse sine function, we get \( \theta = \pi/2 \). \n2. For \( \sin(\theta) + 1 = 0 \), subtracting 1 from both sides gives \( \sin(\theta) = -1 \). Using the inverse sine function, we get \( \theta = 3\pi/2 \).

Step 3: Check the range of \( \theta \)

After solving the equations, we have \( \theta = \pi/2 \) and \( \theta = 3\pi/2 \). Both solutions fall within the required range of [0, \(2\pi\)), hence they are both valid solutions for the original equation.

Key Concepts

Sine Function

The sine function is one of the primary functions in trigonometry, commonly notated as \( \sin \theta \). It's defined based on a right-angle triangle or the unit circle. It measures the ratio of the length of the side opposite the angle (\( \theta \)) to the hypotenuse in a right-angled triangle. When using the unit circle, the sine of an angle corresponds to the y-coordinate of the point on the circle.In our original exercise, we encountered the equation \( (\sin \theta - 1)(\sin \theta + 1) = 0 \). This means that \( \sin (\theta) \) can take values such as -1, 0, or 1. The sine function uniquely cycles every \( 2\pi \) radians, so it will repeat values within this range. Knowing how \( \sin \theta \) behaves within one cycle helps solve trigonometric equations.

Unit Circle

The unit circle is a crucial concept when dealing with trigonometric functions, including the sine function. It is a circle with a radius of 1, centered at the origin (0,0) in the coordinate plane. Angles measured in radians wrap around this circle.For each angle \( \theta \), the coordinates \((x, y)\) on the unit circle are \( (\cos \theta, \sin \theta) \). This makes the unit circle a powerful tool to visualize and solve trigonometric equations, as in the given problem.

• For \( \sin \theta = 1 \), the y-coordinate reaches its maximum, corresponding to \( \theta = \pi/2 \).
• For \( \sin \theta = -1 \), the y-coordinate is at its minimum, corresponding to \( \theta = 3\pi/2 \).

Understanding these positions helps in finding the specific angles that satisfy a given trigonometric equation.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to determine an angle given a trigonometric ratio. For example, the inverse sine function, often notated as \( \sin^{-1} \) or arcsin, helps to find \( \theta \) when we know \( \sin(\theta) \).In the original exercise, the equation \( \sin(\theta) = 1 \) was solved using the inverse sine function or arcsin, resulting in \( \theta = \pi/2 \). Similarly, for \( \sin(\theta) = -1 \), it provided \( \theta = 3\pi/2 \). Inverse functions are limited to principal values to ensure a unique solution:

• The principal range for \( \sin^{-1} \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), but due to periodicity, solutions within \([0, 2\pi)\) can be found by considering symmetrical positions on the unit circle.

By understanding inverse trigonometric functions, you can effectively solve equations where the sine value is specified, identifying all angle solutions within a given range.