Evaluate Limit Of Trigonometric Expression As X Approaches Π/2

Introduction

In this article, we will delve into the evaluation of a specific limit involving trigonometric functions. The problem at hand involves finding the limit of the expression ${\frac{(1 - \sin x)(1 - \sin^2 x) \cdots (1 - \sin^n x)}{(\cos x)^{2n}}}$ as x approaches ${\frac{\pi}{2}}$. This problem is an interesting exercise in applying various techniques of calculus, including L'Hôpital's Rule and trigonometric identities. Understanding this limit requires a solid grasp of calculus concepts and a bit of algebraic manipulation.

Problem Statement

The given limit we need to evaluate is:

${ \lim_{x \to \frac{\pi}{2}} \frac{(1 - \sin x)(1 - \sin^2 x) \cdots (1 - \sin^n x)}{(\cos x)^{2n}} }$

This limit is of the indeterminate form ${\frac{0}{0}}$, which suggests that we might need to employ L'Hôpital's Rule or manipulate the expression to find an equivalent form that is easier to evaluate. The presence of products of trigonometric functions in the numerator and a power of cosine in the denominator adds a layer of complexity to the problem. To effectively tackle this limit, we will need to carefully analyze each component of the expression and identify suitable strategies for simplification.

Preliminary Analysis

Before diving into the solution, let's perform a preliminary analysis of the expression. As x approaches ${\frac{\pi}{2}}$, ${\sin x}$ approaches 1, making each term in the product in the numerator approach 0. Specifically, each term ${(1 - \sin^k x)}$ for ${k = 1, 2, \ldots, n}$ approaches 0. The denominator, ${(\cos x)^{2n}}$, also approaches 0 since ${\cos(\frac{\pi}{2}) = 0}$. This confirms that the limit is indeed of the form ${\frac{0}{0}}$, an indeterminate form. This indeterminate form is a crucial observation, as it guides our approach towards using L'Hôpital's Rule or algebraic manipulation to simplify and evaluate the limit.

The presence of the product of terms in the numerator suggests that we might benefit from considering logarithms or other transformations that can convert the product into a sum. Additionally, we can explore trigonometric identities to rewrite the expression in a more manageable form. For instance, we can rewrite ${(1 - \sin^2 x)}$ as ${\cos^2 x}$, which will help relate the numerator to the denominator. Such manipulations are often key to unraveling complex limits.

Solution Strategy

To evaluate the limit, we'll employ a combination of algebraic manipulation, trigonometric identities, and L'Hôpital's Rule. Here's a step-by-step strategy:

1. Rewrite the numerator: Use the identity ${1 - \sin^2 x = \cos^2 x}$ to simplify the terms in the numerator.
2. Factor and simplify: Look for common factors between the numerator and the denominator and simplify the expression.
3. Apply L'Hôpital's Rule: If the limit is still in an indeterminate form, apply L'Hôpital's Rule by differentiating the numerator and the denominator separately.
4. Evaluate the limit: After applying L'Hôpital's Rule (if necessary) and simplifying, evaluate the limit as x approaches ${\frac{\pi}{2}}$.

By following these steps, we aim to methodically break down the complex limit into smaller, more manageable parts, eventually leading to a clear evaluation.

Detailed Solution

Let's proceed with the solution step by step.

Step 1: Rewrite the numerator

The numerator is given by:

${ (1 - \sin x)(1 - \sin^2 x) \cdots (1 - \sin^n x) }$

We can rewrite each term using the identity ${a^2 - b^2 = (a - b)(a + b)}$ and the trigonometric identity ${1 - \sin^2 x = \cos^2 x}$. Specifically, we can rewrite the terms as follows:

• ${1 - \sin x}$ remains as is.
• ${1 - \sin^2 x = \cos^2 x}$
• ${1 - \sin^3 x = (1 - \sin x)(1 + \sin x + \sin^2 x)}$
• ${1 - \sin^4 x = (1 - \sin^2 x)(1 + \sin^2 x) = \cos^2 x (1 + \sin^2 x)}$

And so on. In general, for any integer ${k}$, we have:

${ 1 - \sin^k x = (1 - \sin x)(1 + \sin x + \sin^2 x + \cdots + \sin^{k-1} x) }$

For even powers, we can use the difference of squares identity. For odd powers, we factor out ${(1 - \sin x)}$.

Step 2: Factor and Simplify

Now, let's rewrite the numerator using the factorizations we derived:

${ \begin{aligned} (1 - \sin x)(1 - \sin^2 x) \cdots (1 - \sin^n x) &= (1 - \sin x)(\cos^2 x)(1 - \sin^3 x) \cdots (1 - \sin^n x) \\ &= (1 - \sin x)(\cos^2 x)(1 - \sin x)(1 + \sin x + \sin^2 x) \cdots (1 - \sin^n x) \end{aligned} }$

We can see that the term ${(1 - \sin x)}$ appears multiple times in the product. Let's count how many times it appears. For each term ${(1 - \sin^k x)}$, where ${k}$ is a positive integer, ${(1 - \sin x)}$ will be a factor if ${k}$ is not 2 or any other even number where we directly have a ${\cos^2 x}$ term. The total power of ${(1 - \sin x)}$ in the numerator is thus more complex to determine directly due to the varying factorizations.

Instead, let's focus on rewriting each term in the numerator using the general factorization:

${ 1 - \sin^k x = (1 - \sin x) \sum_{j=0}^{k-1} \sin^j x }$

Applying this to the numerator, we get:

${ \prod_{k=1}^{n} (1 - \sin^k x) = \prod_{k=1}^{n} \left[ (1 - \sin x) \sum_{j=0}^{k-1} \sin^j x \right] }$

This can be rewritten as:

${ (1 - \sin x)^n \prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right) }$

Step 3: Rewrite the Expression

Now, we can rewrite the entire expression as:

${ \frac{(1 - \sin x)^n \prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right)}{(\cos x)^{2n}} }$

Let's use the identity ${\cos^2 x = 1 - \sin^2 x = (1 - \sin x)(1 + \sin x)}$. Thus, ${\cos x = \sqrt{(1 - \sin x)(1 + \sin x)}}$, and ${(\cos x)^{2n} = (1 - \sin x)^n (1 + \sin x)^n}$. Substituting this into our expression, we get:

${ \frac{(1 - \sin x)^n \prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right)}{(1 - \sin x)^n (1 + \sin x)^n} }$

We can cancel out the ${(1 - \sin x)^n}$ terms:

${ \frac{\prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right)}{(1 + \sin x)^n} }$

Step 4: Evaluate the Limit

Now, let's evaluate the limit as x approaches ${\frac{\pi}{2}}$:

${ \lim_{x \to \frac{\pi}{2}} \frac{\prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right)}{(1 + \sin x)^n} }$

As x approaches ${\frac{\pi}{2}}$, ${\sin x}$ approaches 1. Thus, we have:

${ \lim_{x \to \frac{\pi}{2}} \sum_{j=0}^{k-1} \sin^j x = \sum_{j=0}^{k-1} 1^j = k }$

Therefore, the product in the numerator becomes:

${ \prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right) \to \prod_{k=1}^{n} k = n! }$

And the denominator becomes:

${ (1 + \sin x)^n \to (1 + 1)^n = 2^n }$

Thus, the limit is:

${ \lim_{x \to \frac{\pi}{2}} \frac{\prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right)}{(1 + \sin x)^n} = \frac{n!}{2^n} }$

Conclusion

By carefully applying trigonometric identities, algebraic manipulation, and evaluating limits, we have found that:

${ \lim_{x \to \frac{\pi}{2}} \frac{(1 - \sin x)(1 - \sin^2 x) \cdots (1 - \sin^n x)}{(\cos x)^{2n}} = \frac{n!}{2^n} }$

This problem showcases the importance of combining different mathematical techniques to solve complex problems. The initial indeterminate form required us to rewrite the expression strategically, ultimately leading to a clear and concise result.

Final Answer

The final answer is ${\frac{n!}{2^n}}$.

Evaluating trigonometric limits can be challenging, especially when dealing with indeterminate forms. In this article, we will dissect a particular limit problem: finding the limit of the expression ${\frac{(1 - \sin x)(1 - \sin^2 x) \cdots (1 - \sin^n x)}{(\cos x)^{2n}}}$ as x approaches ${\frac{\pi}{2}}$. This problem is not only an excellent exercise in calculus but also showcases the power of combining different mathematical techniques, such as L'Hôpital's Rule, trigonometric identities, and algebraic manipulation. By understanding and mastering such problems, you can significantly enhance your problem-solving skills in calculus and related fields.

Understanding the Trigonometric Limit Problem

To effectively address the problem, it's crucial to understand the nature of the limit we are trying to evaluate. The given expression is:

${ \lim_{x \to \frac{\pi}{2}} \frac{(1 - \sin x)(1 - \sin^2 x) \cdots (1 - \sin^n x)}{(\cos x)^{2n}} }$

As x approaches ${\frac{\pi}{2}}$, the sine function ${\sin x}$ approaches 1, and the cosine function ${\cos x}$ approaches 0. Consequently, each term in the product in the numerator, specifically ${(1 - \sin^k x)}$ for ${k = 1, 2, \ldots, n}$, tends towards 0. Simultaneously, the denominator, which is ${(\cos x)^{2n}}$, also approaches 0. This situation clearly indicates that we are dealing with an indeterminate form of type ${\frac{0}{0}}$, which is a classic scenario where L'Hôpital's Rule or algebraic simplification techniques can be applied.

The presence of the product in the numerator suggests that we might need to explore methods that can transform the product into a sum, such as logarithmic transformations. Moreover, leveraging trigonometric identities, such as ${1 - \sin^2 x = \cos^2 x}$, can help us establish a connection between the numerator and the denominator, thereby simplifying the expression. Recognizing these aspects is vital for devising an effective strategy to evaluate the limit.

Developing a Solution Strategy for the Trigonometric Limit

Given the complexity of the expression, a systematic approach is essential. Our solution strategy will involve the following key steps:

1. Simplifying the Numerator: Begin by rewriting each term in the numerator to make the expression more manageable. We will use algebraic identities and trigonometric relationships to simplify the terms involving sine functions.
2. Applying Trigonometric Identities: Use the fundamental trigonometric identity ${1 - \sin^2 x = \cos^2 x}$ to simplify the expression further. This identity is crucial for relating the numerator to the denominator.
3. Factoring and Cancellation: Identify common factors between the numerator and the denominator. Canceling out these common factors can significantly simplify the expression, making it easier to evaluate the limit.
4. Evaluating the Limit: After simplifying the expression, evaluate the limit as x approaches ${\frac{\pi}{2}}$. If the expression is still in an indeterminate form, we may need to apply L'Hôpital's Rule.
5. Applying L'Hôpital's Rule (if necessary): If the limit remains indeterminate after the initial simplifications, L'Hôpital's Rule can be used by differentiating the numerator and the denominator separately. This step often helps in resolving indeterminate forms.
6. Final Evaluation: Once the expression is sufficiently simplified, we will evaluate the limit to obtain the final answer.

By adhering to this structured approach, we can systematically dissect the problem and arrive at the solution in a clear and concise manner. Each step is designed to transform the original complex expression into a simpler form that can be easily evaluated.

Detailed Steps to Solve the Trigonometric Limit

Let's now walk through the detailed steps to solve the limit problem.

Step 1: Simplify the Numerator Using Trigonometric Identities

The numerator of the expression is given by:

${ (1 - \sin x)(1 - \sin^2 x) \cdots (1 - \sin^n x) }$

We can use trigonometric identities to rewrite the terms in the numerator. Specifically, the term ${(1 - \sin^2 x)}$ can be rewritten using the identity ${1 - \sin^2 x = \cos^2 x}$. This gives us:

${ (1 - \sin x)(\cos^2 x)(1 - \sin^3 x) \cdots (1 - \sin^n x) }$

For the general term ${(1 - \sin^k x)}$, we can factor it using the identity:

${ 1 - a^k = (1 - a)(1 + a + a^2 + \cdots + a^{k-1}) }$

Thus,

${ 1 - \sin^k x = (1 - \sin x)(1 + \sin x + \sin^2 x + \cdots + \sin^{k-1} x) }$

Step 2: Rewrite the Numerator in a Factored Form

Using the factorization from Step 1, we can rewrite the entire numerator as:

${ \begin{aligned} (1 - \sin x)(1 - \sin^2 x) \cdots (1 - \sin^n x) &= (1 - \sin x)(1 - \sin^2 x)(1 - \sin^3 x) \cdots (1 - \sin^n x) \\ &= (1 - \sin x)(\cos^2 x)(1 - \sin x)(1 + \sin x + \sin^2 x) \cdots (1 - \sin^n x) \end{aligned} }$

We notice that the term ${(1 - \sin x)}$ appears multiple times. To simplify, let’s rewrite each term ${(1 - \sin^k x)}$ using the general factorization:

${ 1 - \sin^k x = (1 - \sin x) \sum_{j=0}^{k-1} \sin^j x }$

Applying this to the numerator gives us:

${ \prod_{k=1}^{n} (1 - \sin^k x) = \prod_{k=1}^{n} \left[ (1 - \sin x) \sum_{j=0}^{k-1} \sin^j x \right] }$

Which can be rewritten as:

${ (1 - \sin x)^n \prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right) }$

Step 3: Rewrite the Entire Expression

Now, we rewrite the entire limit expression:

${ \lim_{x \to \frac{\pi}{2}} \frac{(1 - \sin x)^n \prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right)}{(\cos x)^{2n}} }$

We use the trigonometric identity ${\cos^2 x = 1 - \sin^2 x = (1 - \sin x)(1 + \sin x)}$. Therefore, ${\cos x = \sqrt{(1 - \sin x)(1 + \sin x)}}$, and ${(\cos x)^{2n} = (1 - \sin x)^n (1 + \sin x)^n}$. Substituting this into the expression, we get:

${ \lim_{x \to \frac{\pi}{2}} \frac{(1 - \sin x)^n \prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right)}{(1 - \sin x)^n (1 + \sin x)^n} }$

We can cancel out the ${(1 - \sin x)^n}$ terms:

${ \lim_{x \to \frac{\pi}{2}} \frac{\prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right)}{(1 + \sin x)^n} }$

Step 4: Evaluate the Limit

Now, we evaluate the limit as x approaches ${\frac{\pi}{2}}$. As x approaches ${\frac{\pi}{2}}$, ${\sin x}$ approaches 1. Thus:

${ \lim_{x \to \frac{\pi}{2}} \sum_{j=0}^{k-1} \sin^j x = \sum_{j=0}^{k-1} 1^j = k }$

Therefore, the product in the numerator approaches:

${ \prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right) \to \prod_{k=1}^{n} k = n! }$

And the denominator approaches:

${ (1 + \sin x)^n \to (1 + 1)^n = 2^n }$

Thus, the limit is:

${ \lim_{x \to \frac{\pi}{2}} \frac{\prod_{k=1}^{n} \left( \sum_{j=0}^{k-1} \sin^j x \right)}{(1 + \sin x)^n} = \frac{n!}{2^n} }$

Conclusion: The Final Result of the Trigonometric Limit

Through a series of careful steps involving trigonometric identities, algebraic manipulation, and limit evaluation, we have successfully found that:

${ \lim_{x \to \frac{\pi}{2}} \frac{(1 - \sin x)(1 - \sin^2 x) \cdots (1 - \sin^n x)}{(\cos x)^{2n}} = \frac{n!}{2^n} }$

This problem underscores the importance of a multi-faceted approach when dealing with complex mathematical problems. The initial indeterminate form required us to strategically rewrite the expression, revealing the underlying structure and leading us to the final, elegant result. The final answer is ${\frac{n!}{2^n}}$, which demonstrates how seemingly complex limits can be resolved with the right combination of techniques.

This exploration also highlights the beauty and interconnectedness of mathematical concepts. By understanding and applying these techniques, one can approach a wide array of problems in calculus and related fields with confidence and precision.

Final Answer

The final answer is ${\frac{n!}{2^n}}$.