Question

The maximum value of $\left(\cos {\alpha }_1\right).\left(\cos {\alpha }_2\right)...\left(\cos {\alpha }_n\right),$ under the restrictions $0\le {\alpha }_1,{\alpha }_2,...{\alpha }_n\le \frac{\pi }{2}$ and $\left(\cot {\alpha }_1\right).\left(\cot {\alpha }_2\right)...\left(\cot {\alpha }_n\right)=1$ is :

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

Answer 1

Answer: (A)

Given condition:

$\left(\cot {\alpha }_1\right).\left(\cot {\alpha }_2\right)...\left(\cot {\alpha }_n\right)=1$

$\Rightarrow {\prod }_{i=1}^n\cot {\alpha }_i=1$

$\Rightarrow \frac{{\prod }_{i=1}^n\cos {\alpha }_i}{{\prod }_{i=1}^n\sin {\alpha }_i}=1$

$\Rightarrow {\prod }_{i=1}^n\cos {\alpha }_i={\prod }_{i=1}^n\sin {\alpha }_i$)

$\Rightarrow \cos {\alpha }_1\cos {\alpha }_2\cos {\alpha }_3....\cos {\alpha }_n=\sin {\alpha }_1\sin {\alpha }_2\sin {\alpha }_3.....\sin {\alpha }_n$

This is possible only if ${\alpha }_1={\alpha }_2={\alpha }_3={\alpha }_4....={\alpha }_n=\frac{\pi }{4}$

Now, the maximum value of $\cos {\alpha }_1\cos {\alpha }_2\cos {\alpha }_3....\cos {\alpha }_n$

$=\cos \frac{\pi }{4}\cos \frac{\pi }{4}\cos \frac{\pi }{4}....\cos \frac{\pi }{4}$

$=\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}....n$-times

$=(\frac{1}{\sqrt{2}}{)}^n$

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

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

$\therefore \cos {\alpha }_1\cos {\alpha }_2\cos {\alpha }_3....\cos {\alpha }_n=\frac{1}{2^{\frac{n}{2}}}$

Hence, Option A is correct.