The maximum value of $\left(\cos \alpha_1\right) \cdot\left(\cos \alpha_2\right) \ldots\left(\cos…

The maximum value of $\left(\cos \alpha_1\right) \cdot\left(\cos \alpha_2\right) \ldots\left(\cos \alpha_n\right)$ under the constraints $0 \leq \alpha_1, \alpha_2, \ldots, \alpha_n \leq \frac{\pi}{2}$ and $\left(\cot \alpha_1\right) \cdot\left(\cot \alpha_2\right) \ldots\left(\cot \alpha_n\right)=1$ is
  1. $\frac{1}{2^{\left(\frac{\mathrm{n}}{2}\right)}}$
  2. $\frac{1}{2^{\mathrm{n}}}$
  3. $2^n$
  4. $2^{\frac{n}{2}}$

Solution

$\begin{array}{ll} & \text { Here, }\left(\cot \alpha_1\right)\left(\cot \alpha_2\right) \ldots\left(\cot \alpha_n\right)=1 \\ \therefore \quad & \cos \alpha_1 \cdot \cos \alpha_2 \ldots \cos \alpha_n \\ & =\sin \alpha_1 \cdot \sin \alpha_2 \ldots \sin \alpha_n \\ & \text { Now, }\left(\cos \alpha_1 \cdot \cos \alpha_2 \ldots \cos \alpha_n\right)^2...(i) \end{array}$ $=\left(\cos \alpha_1 \cdot \cos \alpha_2 \ldots \cos \alpha_n\right)$ $\left(\cos \dot{\alpha}_1 \cdot \cos \alpha_2 \ldots \cos \alpha_n\right)$ $=\left(\cos \alpha_1 \cdot \cos \alpha_2 \ldots \cos \alpha_n\right)$ $\begin{array}{r}\left(\sin \alpha_1 \cdot \sin \alpha_2 \ldots \sin \alpha_n\right) \\ \ldots[\text { From }(i)]\end{array}$ $=\frac{1}{2^{\mathrm{n}}} \sin 2 \alpha_1 \cdot \sin 2 \alpha_2 \ldots \sin 2 \alpha_n$ $\ldots[\because \sin 2 \mathrm{~A}=2 \sin \mathrm{~A} \cos \mathrm{~A}]$ But each of $\sin 2 \alpha_i \leq 1$ $\therefore \quad\left(\cos \alpha_1 \cdot \cos \alpha_2 \ldots \cos \alpha_n\right)^2 \leq \frac{1}{2^n}$
But each of $\cos \alpha_i$ is positive $\therefore \quad \cos \alpha_1 \cdot \cos \alpha_2 \ldots \cos \alpha_n \leq \sqrt{\frac{1}{2^n}}=\frac{1}{2^{\frac{n}{2}}}$

Asked in: MHT CET 2024 (16 May Shift 1)

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