Problem 27 - A Complete Resource Book in Mathematics for JEE Main

Question

The value of \(\cos \theta \cdot \cos 2 \theta \cdot \cos 2^{2} \theta \ldots \cos 2^{n-1} \theta\) for \(\theta=\frac{\pi}{2^{n}+1}\) is (A) 1 (B) \(\frac{1}{2^{n}}\) (C) \(2^{n}\) (D) none of these

Step-by-Step Solution

Answer

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

Step 1: Understand the Exercise

We need to evaluate the product of cosines: \( \cos \theta \cdot \cos 2 \theta \cdot \cos 2^2 \theta \cdot \ldots \cdot \cos 2^{n-1} \theta \) where \( \theta = \frac{\pi}{2^n + 1} \). We want to determine its value.

Step 2: Recall the Angle Doubling Property

Observe that \( \theta, 2\theta, 2^2\theta, \dots, 2^{n-1}\theta \) are an arithmetic sequence. Each angle can be found by doubling the previous one.

Step 3: Use the Identity for Composite Angles

To solve this, we can use the identity for the product of cosines: \( \cos \theta \cdot \cos 2\theta \cdot \cos 4\theta \cdots \cos 2^{n-1}\theta = \frac{\sin 2^n \theta}{2^n \sin \theta} \).

Step 4: Substitute Given Values

Substitute \( \theta = \frac{\pi}{2^n + 1} \) in the identity: \( \frac{\sin 2^n \theta}{2^n \sin \theta} = \frac{\sin\left(2^n \cdot \frac{\pi}{2^n + 1}\right)}{2^n \sin\left(\frac{\pi}{2^n + 1}\right)} \).

Step 5: Simplify the Expression

The expression becomes \( \frac{\sin\left(\frac{2^n \pi}{2^n + 1}\right)}{2^n \sin\left(\frac{\pi}{2^n + 1}\right)} \). Since \( \sin(\pi - x) = \sin x \), the numerator \( \sin\left(\frac{2^n \pi}{2^n + 1}\right) = \sin\left(\pi - \frac{\pi}{2^n + 1}\right) = \sin\left(\frac{\pi}{2^n + 1}\right) \).

Step 6: Calculate the Final Value

Thus, the entire expression simplifies to \( \frac{\sin\left(\frac{\pi}{2^n + 1}\right)}{2^n \sin\left(\frac{\pi}{2^n + 1}\right)} = \frac{1}{2^n} \). This is option (B) in the given choices.

Key Concepts

Product of Cosines

When exploring trigonometric identities, the product of cosines plays a crucial role in simplifying and solving complex expressions. For instance, calculating \( \cos \theta \cdot \cos 2\theta \cdot \cos 2^2\theta \cdots \cos 2^{n-1}\theta \) involves systematically decomposing the expression into manageable parts. To start, it's important to identify each term as a cosine of an increasingly doubled angle.

• Begin with \( \cos \theta \).
• Progress to \( \cos 2\theta \), \( \cos 4\theta \), and further up to \( \cos 2^{n-1}\theta \).

Each subsequent term can be seen as a functional progression in a sequence formed by angle doubling.
Understanding this sequential formation provides a foundation that aids in employing cosine product identities for simplification. These identities transform the entire product into a fraction, aiding further computations.

Angle Doubling

Angle doubling is a powerful concept in trigonometry that elicits detailed patterns in angles, especially when working with series or sequences. This concept emerges prominently while evaluating expressions like the product of cosines.Consider the angles \( \theta, 2\theta, 4\theta, \dots, 2^{n-1}\theta \). Each angle is derived by simply doubling its predecessor. This forms a predictable path of quickly increasing angles, a key characteristic of geometric progressions.

• Doubling elicits new angles such as \( 2\theta \) and \( 4\theta \).
• As you progress, the angle quickly multiplies up to \( 2^{n-1}\theta \).

This characteristic is essential in simplifying trigonometric expressions and discovering underlying symmetrical properties inherent in such problems. When coupled with identities, it provides an effective strategy for both understanding and solving problems involving cosine products.

Trigonometric Simplification

Trigonometric simplification aims to reduce complex trigonometric expressions to more manageable, simpler forms. In this problem, it refers to using identities to evaluate the product of cosines. Simplification helps translate complex expressions into neat, solvable forms.
In the context of this exercise, the given expression is simplified using identities related to the product of series of angles. Consider the identity:\[\cos \theta \cdot \cos 2\theta \cdot \cos 4\theta \, \ldots \, \cos 2^{n-1}\theta = \frac{\sin 2^n \theta}{2^n \sin \theta}\]By substituting \( \theta = \frac{\pi}{2^n + 1} \), we simplify the expression to:\[\frac{\sin\left(\frac{2^n \pi}{2^n + 1}\right)}{2^n \sin\left(\frac{\pi}{2^n + 1}\right)}\]Using the identity \( \sin(\pi - x) = \sin x \), the numerator simplifies to equal the denominator. This process reduces the entire expression to \( \frac{1}{2^n} \), which matches the given options.Applying these trigonometric identities efficiently minimizes steps and clarifies the solution path.