Question

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

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

Answer 1

Answer: (A)

We are given that,
$\left(\cot {\alpha }_1\right)\cdot \left(\cot {\alpha }_2\right)\dots \left(\cot {\alpha }_n\right)=1$
$\Rightarrow (\cos {\alpha }_1\cdot \left(\cos {\alpha }_2\right)\dots \left(\cos {\alpha }_n\right)=\left(\sin {\alpha }_1\right)\cdot \left(\sin {\alpha }_2\right)\dots \left(\sin {\alpha }_n\right)...(i)$
Let $y=\left(\cos {\alpha }_1\right)\cdot \left(\cos {\alpha }_2\right)\dots \left(\cos {\alpha }_n\right)$ (to be max)
Squaring both sides, we get
$y^2=\left({\cos }^2{\alpha }_1\right)\cdot \left({\cos }^2{\alpha }_2\right)\dots \left({\cos }^2{\alpha }_n\right)$
$=\cos {\alpha }_1\cdot \sin {\alpha }_1\cdot \cos {\alpha }_2\cdot \sin {\alpha }_2\dots \cos {\alpha }_n\cdot \sin {\alpha }_n$ [using (i)]
$=\frac{1}{2^n}\left[\sin 2{\alpha }_1\cdot \sin 2{\alpha }_2\dots \sin 2{\alpha }_n\right]$
As $0\le {\alpha }_1,{\alpha }_2,\dots {\alpha }_n\le \frac{\pi }{2}$
$\therefore 0\le 2{\alpha }_1,2{\alpha }_2,\dots ,2{\alpha }_n\le \pi$
$\Rightarrow 0\le \sin 2{\alpha }_1,\sin 2{\alpha }_2,\dots \sin 2{\alpha }_n\le 1$
$\therefore y^2\le \frac{1}{2^n}\cdot 1$
$\Rightarrow y\le \frac{1}{2^{n/2}}$
$\therefore$ Maximum value of $y$ is $\frac{1}{2^{n/2}}$