The maximum value of (cos alpha_1) cdot (cos | vedclass

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Question

(A) Given $(\cot \alpha_1)(\cot \alpha_2) \ldots (\cot \alpha_n) = 1$. This implies $(\cos \alpha_1 \cos \alpha_2 \ldots \cos \alpha_n) = (\sin \alpha

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$\frac{1}{2^{(n/2)}}$

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(A) Given $(\cot \alpha_1)(\cot \alpha_2) \ldots (\cot \alpha_n) = 1$. This implies $(\cos \alpha_1 \cos \alpha_2 \ldots \cos \alpha_n) = (\sin \alpha_1 \sin \alpha_2 \ldots \sin \alpha_n)$. Let $P = \cos \alpha_1 \cos \alpha_2 \ldots \cos \alpha_n$. Then $P^2 = (\cos \alpha_1 \ldots \cos \alpha_n)(\sin \alpha_1 \ldots \sin \alpha_n)$. $P^2 = \frac{1}{2^n} (2 \sin \alpha_1 \cos \alpha_1) (2 \sin \alpha_2 \cos \alpha_2) \ldots (2 \sin \alpha_n \cos \alpha_n)$. $P^2 = \frac{1}{2^n} \sin(2\alpha_1) \sin(2\alpha_2) \ldots \sin(2\alpha_n)$. Since $\sin(2\alpha_i) \leq 1$ for all $i$,we have $P^2 \leq \frac{1}{2^n}$. Taking the square root,$P \leq \frac{1}{\sqrt{2^n}} = \frac{1}{2^{(n/2)}}$. The maximum value is $\frac{1}{2^{(n/2)}}$.

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

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