The maximum value of cos alpha_1 cdot cos alp | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $\cot \alpha_1 \cdot \cot \alpha_2 \cdots \cot \alpha_n = 1$. This implies $\frac{\cos \alpha_1}{\sin \alpha_1} \cdot \frac{\cos \alpha_2}{\sin \alpha_2} \cdots \frac{\cos \alpha_n}{\sin \alpha_n} = 1$,so $\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n = \sin \alpha_1 \cdot \sin \alpha_2 \cdots \sin \alpha_n$. 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 $i$,we have $P^2 \le \frac{1}{2^n}$. Taking the square root,$P \le \sqrt{\frac{1}{2^n}} = \frac{1}{2^{n/2}}$. The maximum value is $\frac{1}{2^{n/2}}$.

The maximum value of $\cos \alpha_1 \cdot \cos \alpha_2 \cdots \cos \alpha_n$ 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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