Question

If $(\cot \alpha_1)(\cot \alpha_2) \cdots (\cot \alpha_n) = 1$, with $0<\alpha_1, \alpha_2, \ldots, \alpha_n<\frac{\pi}{2}$, then the maximum value of $(\cos \alpha_1)(\cos \alpha_2) \cdots (\cos \alpha_n)$ is

• A. 1/2n/2
• B. 1/2n
• C. 1/2n
• D. 1

Answer 1

The Correct Option is A

Given: Let \( 0<\alpha_1, \alpha_2, \ldots, \alpha_n<\frac{\pi}{2} \), and \[ \cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = 1 \] We are to find the maximum value of \[ (\cos \alpha_1)(\cos \alpha_2) \cdots (\cos \alpha_n) \] Step 1: Express cotangent in terms of sine and cosine. Recall: \[ \cot \alpha = \frac{\cos \alpha}{\sin \alpha} \] So, \[ \cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = \frac{\cos \alpha_1}{\sin \alpha_1} \cdot \frac{\cos \alpha_2}{\sin \alpha_2} \cdots \frac{\cos \alpha_n}{\sin \alpha_n} = \frac{(\cos \alpha_1)(\cos \alpha_2) \cdots (\cos \alpha_n)}{(\sin \alpha_1)(\sin \alpha_2) \cdots (\sin \alpha_n)} \] Given this product equals 1, we have: \[ \frac{\prod_{i=1}^n \cos \alpha_i}{\prod_{i=1}^n \sin \alpha_i} = 1 \Rightarrow \prod_{i=1}^n \cos \alpha_i = \prod_{i=1}^n \sin \alpha_i \] Let: \[ P = \prod_{i=1}^n \cos \alpha_i \Rightarrow P = \prod_{i=1}^n \sin \alpha_i \Rightarrow P^2 = \left( \prod_{i=1}^n \cos \alpha_i \right)\left( \prod_{i=1}^n \sin \alpha_i \right) \] So: \[ P^2 = \prod_{i=1}^n (\cos \alpha_i \sin \alpha_i) = \prod_{i=1}^n \frac{1}{2} \sin 2\alpha_i \Rightarrow P^2 = \frac{1}{2^n} \prod_{i=1}^n \sin 2\alpha_i \] Thus: \[ P = \sqrt{ \frac{1}{2^n} \prod_{i=1}^n \sin 2\alpha_i } \] Step 2: Maximize the product of sines. Since \( 0<\alpha_i<\frac{\pi}{2} \), the values \( 0<2\alpha_i<\pi \), and \( \sin 2\alpha_i \leq 1 \). Using the inequality of arithmetic and geometric means (AM–GM inequality), the product \( \prod_{i=1}^n \sin 2\alpha_i \) is maximized when all \( \sin 2\alpha_i \) are equal. That happens when all \( \alpha_i \) are equal, i.e.: \[ \alpha_1 = \alpha_2 = \cdots = \alpha_n = \alpha \] Then the cotangent condition becomes: \[ \cot^n \alpha = 1 \Rightarrow \cot \alpha = 1 \Rightarrow \alpha = \frac{\pi}{4} \] Now: \[ \cos \alpha = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \Rightarrow \prod_{i=1}^n \cos \alpha_i = \left( \frac{1}{\sqrt{2}} \right)^n = \frac{1}{2^{n/2}} \] Final Answer: \[ \boxed{ \frac{1}{2^{n/2}} } \]