The maximum value of cos alpha_1 cdot cos alp | vedclass

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Question

(A) Given that $\cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = 1$.This implies $\frac{\cos \alpha_1}{\sin \alpha_1} \cdot \frac{\cos \alpha_

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

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(A) Given that $\cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = 1$.This implies $\frac{\cos \alpha_1}{\sin \alpha_1} \cdot \frac{\cos \alpha_2}{\sin \alpha_2} \cdots \frac{\cos \alpha_n}{\sin \alpha_n} = 1$,which means $\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n = \sin \alpha_1 \cdot \sin \alpha_2 \cdots \sin \alpha_n$.Let $P = \cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n$.Then $P^2 = (\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n) \cdot (\sin \alpha_1 \cdot \sin \alpha_2 \cdots \sin \alpha_n)$.Using the identity $\sin 	heta \cos 	heta = \frac{1}{2} \sin 2	heta$,we get $P^2 = \frac{1}{2^n} \sin 2\alpha_1 \cdot \sin 2\alpha_2 \cdots \sin 2\alpha_n$.Since $\sin 2\alpha_i \le 1$ for all $i$,the maximum value of $P^2$ is $\frac{1}{2^n}$.Therefore,the maximum value of $P$ is $\sqrt{\frac{1}{2^n}} = \frac{1}{2^{n/2}}$.

The maximum value of $\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n$ under the restrictions $0 \le \alpha_1, \alpha_2, \dots, \alpha_n \le \frac{\pi}{2}$ and $\cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = 1$ is

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