Solving 1/2 < Sin(n^2/n) ≤ 1 Integral Values Of N
Introduction
In this comprehensive article, we delve into the intricacies of solving the trigonometric inequality 1/2 < sin(n^2/n) ≤ 1. This problem combines algebraic manipulation with trigonometric understanding, making it an excellent exercise for those seeking to enhance their problem-solving skills in mathematics. Our primary focus will be on determining the integral values of n that satisfy the given inequality. We will break down the problem into manageable parts, explain the underlying concepts, and provide a step-by-step solution. To begin, we will simplify the expression inside the sine function, then determine the range of angles for which the sine function satisfies the inequality, and finally, we will find the integer values of n that fall within this range. This process requires a good understanding of trigonometry and inequalities, and we will provide detailed explanations to ensure clarity. Through this article, we aim to provide a clear, concise, and thorough solution that not only answers the question but also enhances the reader's understanding of related mathematical concepts. This exploration is essential for anyone looking to improve their proficiency in trigonometry and problem-solving in mathematics. Let’s start with the basics, setting the foundation for a deeper understanding of how to approach and solve such trigonometric inequalities.
Simplifying the Inequality
Before we plunge into solving the trigonometric inequality, 1/2 < sin(n^2/n) ≤ 1, it is essential to first simplify the expression within the sine function. The term n^2/n can be simplified to n, provided that n is not equal to zero. This simplification significantly reduces the complexity of the problem, allowing us to deal with sin(n) rather than sin(n^2/n). However, it is crucial to remember that we are making an assumption here: n ≠ 0. We will need to consider this condition when we state our final solution to ensure we don't include any extraneous solutions or miss any valid ones. By simplifying the expression, we transform the original inequality into 1/2 < sin(n) ≤ 1, which is far more manageable. This step is a perfect illustration of how simplifying complex expressions can make them more tractable. This simplification allows us to focus on the core trigonometric aspect of the problem. Now that we have a simpler inequality to work with, the next step is to identify the intervals where the sine function satisfies the given conditions. Understanding the sine function's behavior across different quadrants will be key to finding the solutions. The simplification n^2/n to n is a crucial first step in making the problem accessible. It underscores the importance of algebraic manipulation in solving trigonometric problems.
Determining the Range of Angles
Now that we have simplified the inequality to 1/2 < sin(n) ≤ 1, our next task is to determine the range of angles for which this condition holds true. We are looking for values of n (in radians) where the sine of n is strictly greater than 1/2 and less than or equal to 1. To visualize this, consider the unit circle and the sine function's graphical representation. The sine function, sin(n), corresponds to the y-coordinate of a point on the unit circle. The sine function equals 1 at π/2 (90 degrees), and it equals 1/2 at π/6 (30 degrees) and 5π/6 (150 degrees). Therefore, we need to find the intervals where the y-coordinate on the unit circle is between 1/2 and 1. This occurs in the first and second quadrants. The principal values of n that satisfy sin(n) = 1/2 are π/6 and 5π/6. The sine function reaches its maximum value of 1 at π/2. Consequently, the range of n values for which 1/2 < sin(n) ≤ 1 is given by the interval (π/6, 5π/6], considering the periodic nature of the sine function, we need to include all possible solutions. To do this, we add integer multiples of 2π to these values. Therefore, the general solution for n is given by π/6 + 2kπ < n ≤ π/2 + 2kπ and π/2 + 2kπ ≤ n < 5π/6 + 2kπ, where k is an integer. Understanding the periodicity of the sine function is crucial for finding all possible solutions. These intervals represent all angles n for which the sine function lies between 1/2 and 1, inclusive. This step is critical in narrowing down the potential solutions for n. Now, we move on to finding the specific integer values of n that fall within these ranges.
Finding Integer Values of n
Having established the general intervals for n as π/6 + 2kπ < n ≤ π/2 + 2kπ and π/2 + 2kπ ≤ n < 5π/6 + 2kπ, where k is an integer, our next crucial step is to determine the specific integer values of n that fall within these intervals. This requires a careful examination of each interval for different values of k. Let’s start by considering the case when k = 0. In this instance, the intervals become π/6 < n ≤ π/2 and π/2 ≤ n < 5π/6. Approximating the values, we have π/6 ≈ 0.524, π/2 ≈ 1.571, and 5π/6 ≈ 2.618. Therefore, the intervals for k = 0 are approximately 0.524 < n ≤ 1.571 and 1.571 ≤ n < 2.618. Within these intervals, the integer values of n are 1 and 2. Now, let's consider k = 1. The intervals become π/6 + 2π < n ≤ π/2 + 2π and π/2 + 2π ≤ n < 5π/6 + 2π. Approximating these values, we get π/6 + 2π ≈ 6.807, π/2 + 2π ≈ 7.854, and 5π/6 + 2π ≈ 8.901. So, the intervals for k = 1 are approximately 6.807 < n ≤ 7.854 and 7.854 ≤ n < 8.901. Within these intervals, the integer values of n are 7 and 8. For k = 2, the intervals shift further to the right, and the integer values increase accordingly. However, for negative values of k, the intervals shift to the left, and it is important to check if any integer values fall within those intervals as well. By plugging in negative values for k, it can be observed that no negative integers satisfy the given inequality. Therefore, the integral values of n that satisfy the inequality 1/2 < sin(n) ≤ 1 are 1, 2, 7, and 8. This exhaustive search for integer solutions within the defined intervals ensures we capture all possible valid solutions.
Verification of Solutions
After identifying potential solutions, it's imperative to verify each one to ensure it satisfies the original inequality. In our case, we've found the integral values of n to be 1, 2, 7, and 8. To verify these, we will substitute each value back into the original inequality, 1/2 < sin(n) ≤ 1, and check if the condition holds. For n = 1: We need to check if 1/2 < sin(1) ≤ 1. Using a calculator, sin(1) ≈ 0.841. Since 1/2 < 0.841 ≤ 1, n = 1 is a valid solution. For n = 2: Similarly, we check if 1/2 < sin(2) ≤ 1. Using a calculator, sin(2) ≈ 0.909. Since 1/2 < 0.909 ≤ 1, n = 2 is also a valid solution. For n = 7: We check if 1/2 < sin(7) ≤ 1. Using a calculator, sin(7) ≈ 0.657. Since 1/2 < 0.657 ≤ 1, n = 7 is a valid solution. For n = 8: We check if 1/2 < sin(8) ≤ 1. Using a calculator, sin(8) ≈ 0.989. Since 1/2 < 0.989 ≤ 1, n = 8 is a valid solution as well. This verification process is a critical step in problem-solving, ensuring that the obtained solutions are accurate and valid. By substituting each potential solution back into the original inequality, we can confirm whether it satisfies the given conditions. This not only validates our solutions but also enhances our confidence in the problem-solving process. Through this step-by-step verification, we can confidently state that the integral values of n that satisfy the inequality 1/2 < sin(n) ≤ 1 are indeed 1, 2, 7, and 8.
Conclusion
In conclusion, solving the trigonometric inequality 1/2 < sin(n^2/n) ≤ 1 and determining the integral values of n that satisfy it is a multi-step process that requires a strong grasp of both algebraic manipulation and trigonometric principles. We started by simplifying the inequality, reducing n^2/n to n, which transformed the problem into the more manageable form of 1/2 < sin(n) ≤ 1. This simplification is a critical first step, making the problem more approachable and easier to solve. Next, we identified the range of angles for which the sine function satisfies the inequality. By understanding the behavior of the sine function in the unit circle and considering its periodicity, we determined the general intervals for n. These intervals are crucial for pinpointing the specific values of n that meet the given conditions. Then, we meticulously searched for integer values of n within these intervals. By considering different values of k and plugging them into the general solution, we narrowed down the potential solutions to 1, 2, 7, and 8. This process involved careful approximation and evaluation of trigonometric values, emphasizing the importance of precision in mathematical problem-solving. Finally, we verified each potential solution by substituting it back into the original inequality. This verification step is not just a formality; it's an essential safeguard against errors, ensuring that the solutions are accurate and valid. The successful resolution of this problem underscores the value of a systematic approach, combining simplification, trigonometric understanding, interval analysis, and verification. The integral values of n that satisfy the inequality are 1, 2, 7, and 8. This comprehensive solution not only provides the answer but also elucidates the process, making it a valuable learning experience for anyone seeking to enhance their mathematical problem-solving skills. Mastering these techniques can open doors to solving more complex mathematical problems in the future.
