Problem 19 - Precalculus Mathematics for Calculus

Question

Solve the given equation, and list six specific solutions. $$\sin \theta=\frac{\sqrt{2}}{2}$$

Step-by-Step Solution

Answer

The six solutions are \( \frac{\pi}{4}, \frac{3\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}, -\frac{7\pi}{4}, -\frac{5\pi}{4} \).

Step 1: Understand the equation

The given equation is \( \sin \theta = \frac{\sqrt{2}}{2} \). We need to find the angles \( \theta \) that satisfy this equation. The equation tells us that the sine of some angle \( \theta \) equals \( \frac{\sqrt{2}}{2} \).

Step 2: Identify the basic angles

The basic angles where the sine function equals \( \frac{\sqrt{2}}{2} \) are well-known from trigonometry. These angles are \( \theta = \frac{\pi}{4} \) and \( \theta = \frac{3\pi}{4} \).

Step 3: Find the periodic solutions

The sine function is periodic with period \( 2\pi \), which means if \( \theta \) is a solution, then \( \theta + 2k\pi \) for any integer \( k \) is also a solution. Therefore, the general solutions for \( \theta \) are \( \theta = \frac{\pi}{4} + 2k\pi \) and \( \theta = \frac{3\pi}{4} + 2k\pi \) for any integer \( k \).

Step 4: List six specific solutions

Let's choose 3 values of \( k \), say \( k = 0, 1, -1 \), and substitute them to get specific solutions. For \( k = 0 \), we have \( \theta = \frac{\pi}{4} \) and \( \theta = \frac{3\pi}{4} \). For \( k = 1 \), we have \( \theta = \frac{\pi}{4} + 2\pi = \frac{9\pi}{4} \) and \( \theta = \frac{3\pi}{4} + 2\pi = \frac{11\pi}{4} \). For \( k = -1 \), we have \( \theta = \frac{\pi}{4} - 2\pi = -\frac{7\pi}{4} \) and \( \theta = \frac{3\pi}{4} - 2\pi = -\frac{5\pi}{4} \). These six specific solutions are: \( \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{9\pi}{4}, \frac{11\pi}{4}, -\frac{7\pi}{4}, -\frac{5\pi}{4} \).

Key Concepts

Sine Function

The sine function, often represented as \( \sin \theta \), is one of the primary trigonometric functions critical in mathematics, especially in the study of periodic phenomena. It's a measure of the vertical component of a point on a unit circle. When you hear "sine," think of waves or oscillations, like tuning forks or ocean tides.

A key property of \( \sin \theta \) is that it takes values between -1 and 1, no matter the angle. This is due to its origin in right triangle ratios, where it represents the ratio of the length of the side opposite an angle to the hypotenuse in a right triangle. For a unit circle, this is simply the y-coordinate of a given point.

Special values, like \( \sin \frac{\pi}{4} = \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} \), are important to remember. These correspond to specific angles in standard positions where sine reaches specific known values.

Periodic Solutions

Periodic solutions are a central concept when working with the sine function due to its periodic nature. The sine function is periodic with a period of \( 2\pi \). This means that it repeats its values every \( 2\pi \) radians.

In practical terms, if you find one solution to an equation like \( \sin \theta = \frac{\sqrt{2}}{2} \), all other solutions can be generated by adding or subtracting multiples of \( 2\pi \) from this angle.

To elaborate, given \( \theta = \frac{\pi}{4} \) as a solution, all angles that are of the form \( \frac{\pi}{4} + 2k\pi \) (where \( k \) is any integer) are also solutions. This holds for any initial angle that solves the equation, like \( \frac{3\pi}{4} \) in this context.

Listing specific solutions often involves choosing convenient integers for \( k \) in this general form to give specific angles, such as those listed in the original solution.

Angle Measurement

Understanding angle measurement is fundamental to grasping trigonometric equations. Angles can be measured in degrees or radians, with radians being the standard unit in mathematics.

A full circle is \( 360 \) degrees or \( 2\pi \) radians. This equivalence is crucial when working with trigonometric equations, as many solutions are initially given in radians, and converting between the two can help understand the angle more intuitively for daily applications.

In this problem, the solutions were expressed in radians, such as \( \frac{\pi}{4} \) and \( \frac{3\pi}{4} \). To convert these to degrees, multiply by \( \frac{180}{\pi} \), resulting in 45 degrees and 135 degrees, respectively. Knowing these conversions can make working with trigonometric functions and understanding their implications easier.