Question

The maximum value of $\left(\cos \alpha_{1}\right) \cdot\left(\cos \alpha_{2}\right) \ldots\left(\cos \alpha_{n}\right)$. Under the restrictions $0 \leq \alpha_{1}, \alpha_{2}, \ldots \alpha_{n} \leq \frac{\pi}{2}$ and $\left(\cot \alpha_{1}\right) \cdot\left(\cot \alpha_{2}\right) \ldots\left(\cot \alpha_{n}\right)=1$ is

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

Answer 1

Answer: (A)

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