Question

The maximum values of $\cos {\alpha }_1\cdot \cos {\alpha }_2\cdot ...\cdot \cos {\alpha }_n$, under the restrictions $0\le {\alpha }_1,{\alpha }_2,...,{\alpha }_n\le \frac{\pi }{2}$ and $\cot {\alpha }_1\cdot \cot {\alpha }_2\cdot ...\cot {\alpha }_n=1$, is

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

Answer 1

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

For $\cot {\alpha }_1.\cot {\alpha }_2.\cdots \cot {\alpha }_n=1$

Case 1: If $n$ is even

Let ${\alpha }_1{\cdots \alpha }_{\left(\frac{n}{2}\right)}={\theta }^{\circ }$ and ${\alpha }_{(\frac{n}{2}+1)}\cdots {\alpha }_n={(90-\theta )}^{\circ }$

$\Rightarrow f\left(\theta \right)=(\cot {\alpha }_1.\cot {\alpha }_2\cdots \cot {\alpha }_n)=(\frac{\cos {\alpha }_1}{\sin {\alpha }_1}\cdots \frac{\cos {\alpha }_{\left(\frac{n}{2}\right)}}{\sin {\alpha }_{\left(\frac{n}{2}\right)}}.\frac{\cos {\alpha }_{(\frac{n}{2}+1)}}{\sin {\alpha }_{(\frac{n}{2}+1)}}\cdots \frac{\cos {\alpha }_n}{\sin {\alpha }_n})$

$={\left(\sin \theta \cos \theta \right)}^{\frac{n}{2}}={\left(\frac{\sin 2\theta }{2}\right)}^{\frac{n}{2}}$

$\Rightarrow f\left(\theta \right)$ is maximum when $2\theta ={90}^{\circ }$

$\Rightarrow f\left(\theta \right)={\left(\frac{1}{2}\right)}^{\frac{n}{2}}$

Case 2: When $n$ is odd and ${\alpha }_n={45}^{\circ },{\alpha }_1=\cdots ={\alpha }_{\frac{(n-1)}{2}}={\theta }^{\circ },{\alpha }_{\frac{n+1}{2}}=\cdots ={\alpha }_{n-1}={(90-\theta )}^{\circ }$
$\Rightarrow f\left(\theta \right)={\left(\frac{\sin 2\theta }{2}\right)}^{\frac{(n-1)}{2}}\sin {45}^{\circ }$

Hence, $f\left(\theta \right)$ is maximum when $2\theta ={90}^{\circ }$

$\Rightarrow f\left(\theta \right)={\left(\frac{1}{2}\right)}^{\frac{(n-1)}{2}}.\frac{1}{2^{\frac{1}{2}}}={\left(\frac{1}{2}\right)}^{\frac{n}{2}}$

Case 3: When all ${\alpha }_1={\alpha }_2=\cdots ={\alpha }_n={45}^{\circ }$

$\Rightarrow f\left(\theta \right)={\prod }_{i=1}^n\left(\sin {\alpha }_i\cos {\alpha }_i\right)={\left(\frac{1}{2}\right)}^{\frac{n}{2}}$

Hence, in all 3 cases maximum value of $f\left(\theta \right)={\left(\frac{1}{2}\right)}^{\frac{n}{2}}$