Question

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

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

Answer 1

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

We are given that,

$\Rightarrow$$\left(\cot {\alpha }_1\right)\cdot \left(\cot {\alpha }_2\right)\cdots \left(\cot {\alpha }_n\right)=1$

$\Rightarrow \left(\cos {\alpha }_1\right)\cdot \left(\cos {\alpha }_2\right)\cdots \left(\cos {\alpha }_n\right)=\left(\sin {\alpha }_1\right)\cdot \left(\sin {\alpha }_2\right)\cdots \left(\sin {\alpha }_n\right)$ ...(i)

$\Rightarrow$Let $y=\left(\cos {\alpha }_1\right)\cdot \left(\cos {\alpha }_2\right)\cdots \left(\cos {\alpha }_n\right)$ (to be max.)

Squaring both sides, we get

$\Rightarrow$$y^2=\left({\cos }^2{\alpha }_1\right)\cdot \left({\cos }^2{\alpha }_2\right)\cdots \left({\cos }^2{\alpha }_n\right)$

$\Rightarrow$$y^2=\cos {\alpha }_1\cdot \sin {\alpha }_1\cdot \cos {\alpha }_2\cdot \sin {\alpha }_2\cdots \cos {\alpha }_n\cdot \sin {\alpha }_n$ [using (i)]

$\Rightarrow$$y^2=\frac{1}{2^n}[\sin 2{\alpha }_1\cdot \sin 2{\alpha }_2\cdots \sin {2\alpha }_n]$

As $0\le {\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n\le \frac{\pi }{2}$

Thus $0\le {2\alpha }_1.{2\alpha }_2,\dots ,{2\alpha }_n\le \pi$

$\Rightarrow 0\le \sin 2{\alpha }_1,\sin 2{\alpha }_2,\dots ,\sin 2{\alpha }_n\le 1$

$\Rightarrow y^2\le \frac{1}{2^n}\cdot 1$ $\Rightarrow y\le \frac{1}{2^{n/2}}$

Therefore, maximum value of $y$ is $\frac{1}{2^{n/2}}$.