Question

The maximum value of $\cos {\alpha }_1\cdot \cos {\alpha }_2\cdot \cos {\alpha }_3\cdots \cos {\alpha }_n$ under the restriction
$0\le {\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n\le \frac{\pi }{2}$ and $\cot {\alpha }_1\cdot \cot {\alpha }_2\cdots \cot {\alpha }_n=1$ is

• A. $\frac{n}{2^{\frac{n}{2}}}$
• B. $\frac{1}{2^n}$
• C. $\frac{1}{2^{\frac{n}{2}}}$
• D. $1$

Answer 1

Answer: (C)

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

Given $\left(\cot {\alpha }_1\right)\left(\cot {\alpha }_2\right)\left(\cot {\alpha }_3\right)...\left(\cot {\alpha }_n\right)=1$
$\Rightarrow \prod _{i=1}^n\cos {\alpha }_i=\prod _{i=1}^n\sin {\alpha }_i\Rightarrow \prod _{i=1}^n{\cos }^2{\alpha }_i=\prod _{i=1}^n\sin {\alpha }_i\cos {\alpha }_i$
$\prod _{i=1}^n\sin {\alpha }_i\cos {\alpha }_i=\frac{1}{2^n}\prod _{i=1}^n\sin {2\alpha }_i$

$\Rightarrow \prod _{i=1}^n{\cos }^2{\alpha }_i=\frac{1}{2^n}\prod _{i=1}^n\sin {2\alpha }_i$
So the maximum value of $\prod _{i=1}^n\cos {\alpha }_i$ is $\frac{1}{2^{\frac{n}{2}}}$