The maximum value of $(\cos{\alpha_{1})\ }(\cos{\alpha_{2})\ldots(\cos{\alpha_{n})\ }\ }$under the restriction $0 \leq \alpha_{1,}\alpha_{2},\ldots,\alpha_{n} \leq \frac{\pi}{2}$ and $(\cos{\alpha_{1})}(\cos{\alpha_{2})\ldots(\cos{\alpha_{n}) = 1,\ }\ }$is

Question

Dr. Abubacker · Accepted Answer

(a)
Since, ${(cot}{\alpha_{1})\left( \cot\alpha_{2} ight)\ldots\left( \cot\alpha_{n} ight) = 1}$
$\Rightarrow (\cos{\alpha_{1})(\cos{\alpha_{1})\ldots\left( \cos\alpha_{n} ight) = \left( \sin\alpha_{1} ight)\left( \sin\alpha_{2} ight)\ldots\left( \sin\alpha_{n} ight)}}$
$\Rightarrow \cos^{2}{\alpha_{1\ }\cos^{2}{\alpha_{2\ldots}\cos^{2}\alpha_{n}}} = \frac{\sin{{2\alpha}_{1}\sin{{2\alpha}_{2}\ldots\sin{2\alpha}_{n}}}}{2^{n}}$
$\Rightarrow {\cos\ }{\alpha_{1\ }\cos{\alpha_{2\ldots}\cos\alpha_{n}}} = \left( \frac{\sin{{2\alpha}_{1}\sin{{2\alpha}_{2\ldots}\sin{2\alpha}_{n}}}}{2^{n}} ight)^{1/2}$
Since, maximum value of $\sin{\alpha = 1}$
$	herefore Maximum\ value\ of\cos{\alpha_{1}\ldots\cos{\alpha_{n} = \frac{1}{2^{n/2}}}}$

Q. The maximum value of $(\cos{\alpha_{1})\ }(\cos{\alpha_{2})\ldots(\cos{\alpha_{n})\ }\ }$under the restriction $0 \leq \alpha_{1,}\alpha_{2},\ldots,\alpha_{n} \leq \frac{\pi}{2}$ and $(\cos{\alpha_{1})}(\cos{\alpha_{2})\ldots(\cos{\alpha_{n}) = 1,\ }\ }$is

Explanation

Reviewed by Dr. Abubacker

Want more practice?