Question

The maximum value of $\left(\cos {\alpha }_1\right)\left(\cos {\alpha }_2\right)...\left(\cos {\alpha }_n\right)$ under the restriction $0\le {\alpha }_1,{\alpha }_2,....,{\alpha }_n\le \pi /2$ 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

The correct option is A $\frac{1}{2^{\frac{n}{2}}}$

It is given that
$\cot {\alpha }_1.\cot \left({\alpha }_2\right)...\cot \left({\alpha }_n\right)=1$
Or
$\cos \left({\alpha }_1\right).\cos \left({\alpha }_2\right)..\cos \left({\alpha }_n\right)=\sin \left({\alpha }_1\right).\sin \left({\alpha }_2\right)..\sin \left({\alpha }_n\right)$
$\cos \left({\alpha }_i\right)=\sin \left({\alpha }_i\right)$.
This happens when ${\alpha }_i=\frac{\pi }{4}$ or $\frac{5\pi }{4}$ in $[0,2\pi ]$.
Now for $\cos \left({\alpha }_1\right).\cos \left({\alpha }_2\right)..\cos \left({\alpha }_n\right)$ to be maximum, ${\alpha }_1={\alpha }_2={\alpha }_3=..{\alpha }_n={\alpha }_i$.
It is also given that $0<{\alpha }_i<\frac{\pi }{2}$.
Hence ${\alpha }_i={\alpha }_1={\alpha }_2=..{\alpha }_n=\frac{\pi }{4}$
Hence
$\cos \left({\alpha }_1\right).\cos \left({\alpha }_2\right)..\cos \left({\alpha }_n\right)$
$={\frac{1}{\sqrt{2}}}^n$

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