Problem 14 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The maximum value of \(\left(\cos \alpha_{1}\right)\left(\cos \alpha_{2}\right) \ldots .(\cos a n)\) under the restrictions \(0 \leq \alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \leq \frac{\pi}{2}\) and (cot \(\left.\alpha_{1}\right)\left(\cot \alpha_{2}\right) \ldots(\cot \alpha)=1\) is (A) \(\frac{1}{2^{\pi / 2}}\) (B) \(\frac{1}{2^{x}}\) (C) \(\frac{1}{2 n}\) (D) 1

Step-by-Step Solution

Answer

The maximum value is (B) \(\frac{1}{2^n}\).

Step 1: Understanding the Problem

We are tasked with finding the maximum value of the product \((\cos \alpha_{1})(\cos \alpha_{2})\ldots(\cos \alpha_{n})\) given the angular restrictions and the constraint \((\cot \alpha_{1})(\cot \alpha_{2}) \ldots (\cot \alpha_{n}) = 1\).

Step 2: Rewriting the Trigonometric Identities

Recall that \(\cot \alpha = \frac{\cos \alpha}{\sin \alpha}\). Given the constraint \((\cot \alpha_{1})(\cot \alpha_{2}) \ldots (\cot \alpha_{n}) = 1\), we have \(\frac{(\cos \alpha_{1})(\cos \alpha_{2}) \ldots (\cos \alpha_{n})}{(\sin \alpha_{1})(\sin \alpha_{2}) \ldots (\sin \alpha_{n})} = 1\).

Step 3: Simplifying with the Constraint

The constraint implies \((\cos \alpha_{1})(\cos \alpha_{2}) \ldots (\cos \alpha_{n}) = (\sin \alpha_{1})(\sin \alpha_{2}) \ldots (\sin \alpha_{n})\). Therefore, \((\cos \alpha_{i})^2 = (\sin \alpha_{i})^2\) for each \(i\).

Step 4: Optimizing the Product

To maximize \((\cos \alpha_{1})(\cos \alpha_{2})\ldots(\cos \alpha_{n})\), note that this occurs when each term is as large as possible. Thus, for the maximum, set \(\cos \alpha_{i} = \sin \alpha_{i}\) which occurs when \(\alpha_{i} = \frac{\pi}{4}\).

Step 5: Mathematical Calculation of the Maximum Value

If each \(\alpha_{i} = \frac{\pi}{4}\), then each \(\cos \alpha_{i} = \sin \alpha_{i} = \frac{\sqrt{2}}{2}\). So, \((\cos \alpha_{i})(\cos \alpha_{i}) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}\). Repeated \(n\) times, the product is \(\left(\frac{1}{2}\right)^n = \frac{1}{2^n}\).

Step 6: Final Verification

Verify that this achieved product satisfies the constraint. Since each \(\alpha_i = \frac{\pi}{4}\), the entire condition \((\cot \alpha_{1})(\cot \alpha_{2}) \ldots (\cot \alpha_{n}) = 1\) holds true because \(\cot \frac{\pi}{4} = 1\). Hence, \(1^n = 1\).

Key Concepts

Product of Cosines

The product of cosines is a crucial concept often encountered when dealing with trigonometric identities, especially in solving problems involving multiple angles. In this problem, we are tasked with finding the maximum value of the product \((\cos \alpha_{1})(\cos \alpha_{2})\ldots(\cos \alpha_{n})\). This is under specific constraints where each angle \(\alpha_i\) is between 0 and \(\frac{\pi}{2}\), and all together satisfy the cotangent condition given in the exercise.
The product of cosines can be thought of as multiplying several cosine values together. To maximize this product, each individual cosine should be as large as possible within its restrictions. When angles range from 0 to \(\frac{\pi}{2}\), the cosine function starts from 1 and gradually decreases to 0. Therefore, maximizing each cosine value means picking the angle where cosine isn't too low. In the solution, it is identified that setting each \(\alpha_i = \frac{\pi}{4}\) maximizes the cosine because \(\cos \frac{\pi}{4}\) is \(\frac{\sqrt{2}}{2}\), which is relatively high. Repeating this \(n\) times results in \(\left(\frac{1}{2}\right)^n\) as the maximum product.
This optimal choice not only satisfies the trigonometric identity constraints but also ensures the product of cosines is maximized.

Cubic Equation

Although the exercise does not explicitly discuss a cubic equation, understanding the principles behind solving equations is meaningful when working through the problem. Cubic equations are polynomial equations of degree three, usually expressed in the form \(ax^3 + bx^2 + cx + d = 0\).
Solving cubic equations often involves finding one or more real roots that satisfy the equation. General methods include factoring, using the cubic formula, or approximating solutions using numerical methods. In trigonometric contexts, these ideas can intersect when trying to solve angles for symmetrical properties or balance conditions between product terms.
By applying similar principles of equation-solving to trigonometric identities, students can understand that identifying symmetries, such as recognizing how \(\sin\) and \(\cos\) balance each other out in this problem, is similar to seeking roots that solve a cubic equation. This conceptual bridge helps students build deeper connections across different areas of math.

Cotangent

Cotangent is another fundamental trigonometric function defined as the reciprocal of the tangent function. It can be represented with the formula \(\cot \alpha = \frac{\cos \alpha}{\sin \alpha}\). In this exercise, the cotangent function helps frame the main constraint: \((\cot \alpha_{1})(\cot \alpha_{2})\ldots(\cot \alpha_{n}) = 1\).
This condition implies a balance of sine and cosine across all the angles. Since \(\cot \alpha_{i} = 1\) for all \(\alpha_{i} = \frac{\pi}{4}\) (because \(\tan \frac{\pi}{4}\) is 1 and the reciprocal is also 1), choosing \(\alpha_i\) at these specific values not only maximizes the cosine product but satisfies this cotangent condition neatly.
Therefore, understanding cotangent's role is critical: it forces the solution to find a perfect equilibrium between sine and cosine across all angles, ensuring the condition holds true. By effectively balancing sine and cosine through cotangent, students can approach trigonometric identity problems with greater insight and clarity.