Solving 2sin(θ) = √2: Find Solutions [0, 2π]

by Alex Johnson 45 views

Introduction

Let's dive into the world of trigonometry! In this article, we're going to tackle the equation 2sin(θ) = √2 and find all the solutions for θ within the interval 0 ≤ θ < 2π. This is a classic problem in trigonometry that involves understanding the unit circle, trigonometric functions, and their periodic nature. Whether you're a student brushing up on your math skills or just someone curious about trigonometry, this guide will walk you through each step in a clear and easy-to-understand way. So, grab your pencils and let's get started!

Understanding the Problem

Before we jump into solving the equation, let's make sure we understand what the question is asking. We have the trigonometric equation 2sin(θ) = √2, and we need to find all values of θ that make this equation true. However, we're not looking for any solution; we're only interested in solutions that fall within the interval 0 ≤ θ < 2π. This means we're considering angles from 0 radians up to, but not including, 2π radians, which is one full revolution around the unit circle. Understanding this interval is crucial because the sine function is periodic, meaning it repeats its values at regular intervals. This periodicity will lead to multiple solutions within our specified interval. By limiting our search to 0 ≤ θ < 2π, we're focusing on the solutions within a single period of the sine function.

Breaking Down the Equation

The equation 2sin(θ) = √2 involves the sine function, which relates an angle θ to the ratio of the opposite side to the hypotenuse in a right-angled triangle. In the context of the unit circle, sin(θ) represents the y-coordinate of a point on the circle that corresponds to the angle θ. The unit circle is a circle with a radius of 1 centered at the origin of a coordinate plane. It's a fundamental tool in trigonometry because it allows us to visualize the values of trigonometric functions for different angles. The equation is set up such that twice the sine of θ equals the square root of 2. Our goal is to isolate sin(θ) and then determine which angles have a sine value that satisfies the equation. We can do this by dividing both sides of the equation by 2, which will give us a simpler form to work with: sin(θ) = √2 / 2. This simplified equation tells us that we are looking for angles θ where the y-coordinate on the unit circle is equal to √2 / 2.

Step-by-Step Solution

Now that we have a solid understanding of the problem, let's break down the solution into manageable steps.

Step 1: Isolate sin(θ)

The first step in solving the equation 2sin(θ) = √2 is to isolate the sine function. This means we want to get sin(θ) by itself on one side of the equation. To do this, we divide both sides of the equation by 2:

2sin(θ) / 2 = √2 / 2

This simplifies to:

sin(θ) = √2 / 2

This equation tells us that we're looking for angles θ whose sine value is equal to √2 / 2. This is a significant simplification because it allows us to focus on the values of θ that satisfy this specific condition. We've now transformed the original equation into a more manageable form that we can use to find the solutions.

Step 2: Identify Reference Angles

The next step is to identify the reference angles. A reference angle is the acute angle formed between the terminal side of an angle and the x-axis. It helps us find angles in other quadrants that have the same trigonometric value (in this case, sine value) as our initial angle. We know that sin(θ) = √2 / 2. From our knowledge of the unit circle or trigonometric values of special angles, we know that sin(π/4) = √2 / 2. Therefore, π/4 is our reference angle. The angle π/4 (which is 45 degrees) is a crucial reference because it's a standard angle in trigonometry, and its sine value is a well-known quantity. By identifying this reference angle, we've found one solution to our equation, but we also have a key to finding other solutions in different quadrants.

Step 3: Determine Quadrants

Now that we have our reference angle, we need to determine the quadrants where the sine function is positive because we are looking for sin(θ) = √2 / 2, which is a positive value. Recall that sine corresponds to the y-coordinate on the unit circle. The y-coordinate is positive in the first and second quadrants. Therefore, we need to find angles in the first and second quadrants that have a reference angle of π/4. Understanding the quadrants where trigonometric functions are positive or negative is essential for finding all solutions to trigonometric equations. The acronym "ASTC" (All Students Take Calculus) can be helpful in remembering which functions are positive in each quadrant: All trigonometric functions are positive in the First quadrant, Sine is positive in the Second quadrant, Tangent is positive in the Third quadrant, and Cosine is positive in the Fourth quadrant. In our case, since we're dealing with a positive sine value, we focus on the first and second quadrants.

Step 4: Find Solutions in Each Quadrant

Now, let's find the solutions in the first and second quadrants.

  • Quadrant I: In the first quadrant, the angle is simply the reference angle itself. So, one solution is θ = π/4. This is because the first quadrant is the starting point on the unit circle, and angles in this quadrant are measured directly from the positive x-axis. The sine value at π/4 is √2 / 2, which satisfies our equation. The solution in the first quadrant is straightforward and serves as a base for finding other solutions.

  • Quadrant II: In the second quadrant, to find the angle with the same reference angle, we subtract the reference angle from π (since π represents the angle on the negative x-axis). So, the other solution is θ = π - π/4 = 3π/4. In the second quadrant, angles are measured from the positive x-axis, moving counterclockwise, up to π radians (180 degrees). Subtracting the reference angle from π gives us the angle in the second quadrant that has the same sine value as the reference angle. This angle, 3π/4, is another solution to our equation because its sine value is also √2 / 2.

Step 5: Check the Interval

Finally, we need to make sure that our solutions fall within the given interval 0 ≤ θ < 2π. Both of our solutions, π/4 and 3π/4, are within this interval. Therefore, these are the only solutions to the equation 2sin(θ) = √2 in the interval 0 ≤ θ < 2π. It's crucial to check the interval because trigonometric functions are periodic, and there can be infinitely many solutions if we don't restrict the range. By confirming that our solutions are within the specified interval, we ensure that we've found all the solutions that are relevant to the problem.

Solutions

Therefore, the solutions to the equation 2sin(θ) = √2 on the interval 0 ≤ θ < 2π are:

  • θ = π/4
  • θ = 3π/4

These are the two angles within the interval 0 ≤ θ < 2π where the sine function equals √2 / 2. We've systematically found these solutions by isolating the sine function, identifying the reference angle, determining the quadrants where sine is positive, finding the angles in those quadrants, and verifying that our solutions are within the given interval. This step-by-step approach ensures that we've found all the solutions and that they are correct.

Conclusion

In this article, we've successfully found all solutions to the equation 2sin(θ) = √2 on the interval 0 ≤ θ < 2π. We achieved this by understanding the properties of the sine function, using the unit circle, and applying a systematic approach to problem-solving. Remember, the key to tackling trigonometric equations is to break them down into smaller steps, understand the underlying concepts, and practice consistently. We started by isolating sin(θ), then identified the reference angle, determined the relevant quadrants, found the solutions in each quadrant, and finally, checked that our solutions fell within the given interval. This method can be applied to solving a variety of trigonometric equations.

Trigonometry can seem daunting at first, but with practice and a clear understanding of the concepts, you can master it. Keep exploring, keep practicing, and don't be afraid to ask questions. Understanding trigonometry opens doors to many fields in mathematics and science, so it's a valuable skill to develop. If you want to further enhance your knowledge about trigonometric functions and equations, visit Khan Academy's Trigonometry Section for more detailed explanations and practice problems.