Question

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

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

Answer 1

Answer: (A)

$\text{ Given }\cos {\alpha }_1\cos {\alpha }_2-\cos {\alpha }_3\dots .\cos {\alpha }_n=\sin {\alpha }_1\sin {\alpha }_2\dots \dots \sin {\alpha }_n$
$y=\left(\cos {\alpha }_1\right)\left(\cos {\alpha }_2\right)\dots \dots \left(\cos {\alpha }_n\right)\left(\sin {\alpha }_1\right)\left(\sin {\alpha }_2\right)\left(\sin {\alpha }_3\right)\dots \left(\sin {\alpha }_n\right)$
$[M-1]y^2=\left(\cos {\alpha }_1\right)\left(\cos {\alpha }_2\right)\dots \left(\cos {\alpha }_n\right)\left(\sin {\alpha }_1\right)\left(\sin {\alpha }_2\right)\dots \left(\sin {\alpha }_n\right)$
$y^2=\frac{1}{2^n}\left(\sin 2{\alpha }_1\right)\left(\sin 2{\alpha }_1\right)\dots \left(\sin +{\alpha }_n\right)$
$\therefore y_{\max }^2=\frac{1}{2^n}\Rightarrow y_{\max }=\frac{1}{2^{4/2}}$
$[\text{ M-2] }\frac{1+{\tan }^2{\alpha }_1}{2}\ge {\left(1+{\tan }^2\alpha \right)}^{1/2}=\tan {\alpha }_1$
$\therefore 1+{\tan }^2{\alpha }_1\ge 2\tan {\alpha }_1\sec {\alpha }_1\sin {\alpha }_2\dots .\sin {\alpha }_n\ge 2^{4/2}$
${\sec }^2{\alpha }_1\ge 2\tan {\alpha }_1$
${\sec }^2{\alpha }^2\ge 2\tan {\alpha }_2$
${\sec }^2{\alpha }_n\ge 2\tan {\alpha }_n$
$\therefore {\cos }_1\cos {\alpha }_2\dots \dots \cos {\alpha }_n\le \frac{1}{2^{n/2}}$