Analysis Of The Function Y= (sin 2x / (1 + Cos 2x))^2

In this comprehensive exploration, we delve into the fascinating world of mathematical functions, specifically focusing on the function $y=\left(\frac{\sin 2x}{1+\cos 2x}\right)^2$. This function, while seemingly complex at first glance, reveals a wealth of mathematical properties and behaviors when analyzed through the lens of trigonometry, calculus, and graphical representation. Our discussion will encompass a detailed examination of its domain, range, periodicity, symmetry, critical points, and graphical characteristics. By dissecting each aspect, we aim to provide a thorough understanding of this function's unique features and its place within the broader landscape of mathematical functions.

Unveiling the Domain and Range

Understanding the domain and range is crucial for comprehending the behavior of any function. The domain of a function refers to the set of all possible input values (x-values) for which the function is defined, while the range represents the set of all possible output values (y-values) that the function can produce. For the function $y=\left(\frac{\sin 2x}{1+\cos 2x}\right)^2$, we must consider the restrictions imposed by the trigonometric functions involved and the potential for division by zero.

The presence of the term $\frac{\sin 2x}{1+\cos 2x}$ immediately raises a concern about the denominator. Division by zero is undefined in mathematics, so we must identify any values of x that make $1 + \cos 2x = 0$. This occurs when $\cos 2x = -1$. The cosine function equals -1 at odd multiples of π, meaning $2x = (2n+1)\pi$, where n is an integer. Solving for x, we get $x = \frac{(2n+1)\pi}{2}$. Therefore, the domain of the function excludes these values. We can express the domain as all real numbers x except for $x = \frac{(2n+1)\pi}{2}$, where n is an integer.

To determine the range, we need to analyze the possible values of the function's output. The function is squared, which means the output will always be non-negative. The term $\sin 2x$ oscillates between -1 and 1, and $\cos 2x$ also oscillates between -1 and 1. The denominator $1 + \cos 2x$ will vary between 0 and 2. When $\cos 2x$ approaches -1, the denominator approaches 0, and the fraction $\frac{\sin 2x}{1+\cos 2x}$ can become arbitrarily large (positive or negative). However, since the entire expression is squared, the output will always be non-negative. To find the maximum value, we can analyze the function's behavior as x varies. We'll find that the maximum value of the function is 1. Therefore, the range of the function is $0 \leq y < \infty$.

Periodicity and Symmetry: Unveiling the Function's Rhythmic Patterns

Periodicity and symmetry are fundamental properties that help us understand the repeating patterns and reflective characteristics of a function. A periodic function repeats its values at regular intervals, while symmetry describes how the function's graph behaves under certain transformations, such as reflection or rotation. Analyzing these properties provides valuable insights into the function's overall behavior and structure.

To determine the periodicity of $y=\left(\frac{\sin 2x}{1+\cos 2x}\right)^2$, we need to investigate whether there exists a constant P such that $f(x+P) = f(x)$ for all x in the domain. The trigonometric functions $\sin 2x$ and $\cos 2x$ have a period of π, since their arguments are 2x. This suggests that the period of the entire function might also be π. Let's verify this:

$f(x + \pi) = \left(\frac{\sin 2(x + \pi)}{1 + \cos 2(x + \pi)}\right)^2 = \left(\frac{\sin (2x + 2\pi)}{1 + \cos (2x + 2\pi)}\right)^2$

Since $\sin(2x + 2\pi) = \sin 2x$ and $\cos(2x + 2\pi) = \cos 2x$, we have:

$f(x + \pi) = \left(\frac{\sin 2x}{1 + \cos 2x}\right)^2 = f(x)$

This confirms that the function has a period of π. The function repeats its pattern every π units along the x-axis.

Now, let's examine the symmetry of the function. A function is said to be even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$. To test for even symmetry, we evaluate $f(-x)$:

$f(-x) = \left(\frac{\sin (-2x)}{1 + \cos (-2x)}\right)^2$

Using the properties that $\sin(-2x) = -\sin 2x$ and $\cos(-2x) = \cos 2x$, we get:

$f(-x) = \left(\frac{-\sin 2x}{1 + \cos 2x}\right)^2 = \left(\frac{\sin 2x}{1 + \cos 2x}\right)^2 = f(x)$

Since $f(-x) = f(x)$, the function is even. This means that the graph of the function is symmetric with respect to the y-axis. If we were to fold the graph along the y-axis, the two halves would perfectly overlap.

Identifying Critical Points and Extrema

Critical points are the points where the derivative of a function is either zero or undefined. These points are crucial for identifying local maxima, local minima, and inflection points, which help us understand the function's behavior and sketch its graph accurately. To find the critical points of $y=\left(\frac{\sin 2x}{1+\cos 2x}\right)^2$, we need to compute its derivative and analyze where it equals zero or is undefined.

Let's first simplify the function using trigonometric identities. Recall the identity $\tan x = \frac{\sin x}{\cos x}$. We can rewrite $\sin 2x$ as $2 \sin x \cos x$ and $\cos 2x$ as $2 \cos^2 x - 1$. Substituting these into the function, we get:

$y = \left(\frac{2 \sin x \cos x}{1 + (2 \cos^2 x - 1)}\right)^2 = \left(\frac{2 \sin x \cos x}{2 \cos^2 x}\right)^2 = \left(\frac{\sin x}{\cos x}\right)^2 = \tan^2 x$

Now, the function is simplified to $y = \tan^2 x$, which is much easier to differentiate. The derivative of $\tan x$ is $\sec^2 x$, so we can use the chain rule to find the derivative of $y = \tan^2 x$:

$\frac{dy}{dx} = 2 \tan x \cdot \sec^2 x$

To find the critical points, we need to solve for where $\frac{dy}{dx} = 0$ or is undefined. The derivative is zero when $2 \tan x \sec^2 x = 0$. Since $\sec^2 x$ is always positive (except where cosine is zero, which we'll address shortly), the derivative is zero when $\tan x = 0$. The tangent function is zero at integer multiples of π, so $x = n\pi$, where n is an integer. These are potential critical points.

The derivative is undefined when $\sec^2 x$ is undefined, which occurs when $\cos x = 0$. This happens at $x = \frac{(2n+1)\pi}{2}$, where n is an integer. However, these values are not in the domain of the original function, as they make the denominator $1 + \cos 2x$ equal to zero. Therefore, these points are not critical points in the traditional sense, but they do represent points where the function has vertical asymptotes.

Now, let's analyze the nature of the critical points at $x = n\pi$. We can use the first derivative test or the second derivative test to determine whether these points correspond to local minima, local maxima, or saddle points. The second derivative test involves finding the second derivative of the function:

First, recall that $\frac{dy}{dx} = 2 \tan x \sec^2 x$. We can rewrite this as $\frac{dy}{dx} = 2 \frac{\sin x}{\cos x} \frac{1}{\cos^2 x} = 2 \frac{\sin x}{\cos^3 x}$. Now, we find the second derivative using the quotient rule:

$\frac{d^2y}{dx^2} = 2 \frac{\cos^3 x(\cos x) - \sin x(3 \cos^2 x(-\sin x))}{\cos^6 x} = 2 \frac{\cos^4 x + 3 \sin^2 x \cos^2 x}{\cos^6 x}$

$\frac{d^2y}{dx^2} = 2 \frac{\cos^2 x + 3 \sin^2 x}{\cos^4 x}$

At the critical points $x = n\pi$, $\sin x = 0$ and $\cos x = \pm 1$. Therefore, the second derivative at these points is:

$\frac{d^2y}{dx^2}|_{x=n\pi} = 2 \frac{(\pm 1)^2 + 3(0)^2}{(\pm 1)^4} = 2$

Since the second derivative is positive at these points, the critical points $x = n\pi$ correspond to local minima. The value of the function at these points is $y = \tan^2(n\pi) = 0$. Thus, the function has local minima at $(n\pi, 0)$, where n is an integer.

Graphical Representation: Visualizing the Function's Behavior

Graphing a function provides a visual representation of its behavior, allowing us to observe its key characteristics, such as its shape, intercepts, asymptotes, and extrema. By plotting the function $y=\left(\frac{\sin 2x}{1+\cos 2x}\right)^2$, we can consolidate our understanding of its properties and gain further insights into its overall nature.

Based on our previous analysis, we know the following:

• The domain of the function is all real numbers except $x = \frac{(2n+1)\pi}{2}$, where n is an integer. This means the function has vertical asymptotes at these values.
• The range of the function is $0 \leq y < \infty$. The function's output is always non-negative.
• The function is periodic with a period of π. The pattern repeats every π units along the x-axis.
• The function is even, meaning it is symmetric with respect to the y-axis.
• The function has local minima at $(n\pi, 0)$, where n is an integer.

To sketch the graph, we can start by drawing the vertical asymptotes at $x = \frac{(2n+1)\pi}{2}$. These lines will act as boundaries that the graph approaches but never touches. Then, we can plot the local minima at $(n\pi, 0)$. Since the function is periodic, we can focus on graphing one period, say between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and then repeat the pattern.

Between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, the function has a local minimum at (0, 0). As x approaches $\frac{\pi}{2}$ from the left or $-\frac{\pi}{2}$ from the right, the function's value increases and approaches infinity, as dictated by the vertical asymptotes. Due to the even symmetry, the graph on both sides of the y-axis will be mirror images of each other.

The graph will consist of a series of U-shaped curves, each centered around a multiple of π on the x-axis, touching the x-axis at the minima, and extending upwards towards the vertical asymptotes. The function's values will always be non-negative, and the pattern will repeat every π units.

Using a graphing calculator or software, we can visualize the function and confirm our analysis. The graph will clearly show the periodicity, symmetry, asymptotes, and minima, providing a comprehensive visual understanding of the function's behavior.

Conclusion: A Comprehensive Understanding of $y=(\frac{\sin 2x}{1+\cos 2x})^2$

In conclusion, our detailed analysis of the function $y=\left(\frac{\sin 2x}{1+\cos 2x}\right)^2$ has revealed its multifaceted nature. We began by identifying its domain and range, considering the restrictions imposed by the trigonometric functions and the potential for division by zero. We then explored its periodicity and symmetry, uncovering its repeating patterns and reflective characteristics. By computing the derivative and analyzing critical points, we located local minima and understood the function's behavior near these points. Finally, we discussed the graphical representation, which provided a visual confirmation of our findings.

Through this comprehensive exploration, we have gained a deep understanding of the function's properties and behavior. This analysis serves as a valuable example of how mathematical tools and techniques can be applied to dissect and interpret complex functions, enhancing our mathematical knowledge and problem-solving abilities. The insights gained from this study can be applied to analyze other trigonometric functions and mathematical expressions, further expanding our understanding of the mathematical landscape.

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