Question

The maximum value of \((\cos \alpha_1) \cdot (\cos \alpha_2) \cdots (\cos \alpha_n)\) under the restrictions \(0 \leq \alpha_1, \alpha_2, \ldots, \alpha_n \leq \frac{\pi}{2}\) and \((\cot \alpha_1) \cdot (\cot \alpha_2) \cdots (\cot \alpha_n) = 1\) is

• A. \(\frac{1}{2^{n/2}}\)
• B. \(\frac{1}{2^n}\)
• C. \(\frac{1}{2n}\)
• D. 1

Hint:

Use \(\cos \alpha \sin \alpha = \frac{1}{2} \sin 2\alpha \leq \frac{1}{2}\).

Answer 1

The Correct Option is A

Step 1: Formula / Definition}
\[ \prod_{i=1}^n \cot \alpha_i = 1 \Rightarrow \prod_{i=1}^n \cos \alpha_i = \prod_{i=1}^n \sin \alpha_i \]
Step 2: Calculation / Simplification}
Let \(P = \prod_{i=1}^n \cos \alpha_i\). Then \(P^2 = \prod_{i=1}^n (\sin \alpha_i \cos \alpha_i) = \frac{1}{2^n} \prod_{i=1}^n \sin 2\alpha_i\)
Since \(0 \leq 2\alpha_i \leq \pi\), \(0 \leq \sin 2\alpha_i \leq 1\)
Maximum occurs when \(\sin 2\alpha_i = 1\) for all \(i \Rightarrow 2\alpha_i = \frac{\pi}{2} \Rightarrow \alpha_i = \frac{\pi}{4}\)
\(P^2 \leq \frac{1}{2^n} \Rightarrow P \leq \frac{1}{2^{n/2}}\)
Step 3: Final Answer
\[ \frac{1}{2^{n/2}} \]