Question

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

• A.
• B.
• C.
• D. 1

Answer 1

The correct option is A

(a) Here (cot ${\alpha }_1$).(cot ${\alpha }_2$)....(cot ${\alpha }_n$)= 1
$\therefore$ cos${\alpha }_1$.cos${\alpha }_2$........cos${\alpha }_n$ = sin${\alpha }_1$.sin${\alpha }_2$........sin${\alpha }_n$
Now, (cos${\alpha }_1$.cos${\alpha }_2$........cos${\alpha }_n$)${}^2$
= (cos${\alpha }_1$.cos${\alpha }_2$........cos${\alpha }_n$) (cos${\alpha }_1$.cos${\alpha }_2$........cos${\alpha }_n$)
= (cos${\alpha }_1$.cos${\alpha }_2$........cos${\alpha }_n$) (sin${\alpha }_1$.sin${\alpha }_2$........sin${\alpha }_n$)
= $\frac{1}{2^n}$ sin 2${\alpha }_1$.sin 2${\alpha }_2$........sin 2${\alpha }_n$
But each of sin 2${\alpha }_i$ $\le$ 1
(cos${\alpha }_1$.cos${\alpha }_2$........cos${\alpha }_n$)${}^2$ $\le$ $\frac{1}{2^n}$
But each of cos ${\alpha }_i$, is positive.
$\therefore$ cos${\alpha }_1$.cos${\alpha }_2$........cos${\alpha }_n$ $\le$ $\sqrt{\frac{1}{2^n}}$ = $\frac{1}{2^{n/2}}$.