Trigonometry & Inverse Trigonometry Question - None

Question

The maximum value of $\cos a_1 \cos a_2 \cdots \cos a_n$ under the restriction $0 < a_1, a_2, \ldots, a_n < \frac{\pi}{2}$ and $\cot a_1 \cot a_2 \cdots \cot a_n = 1$ is

Accepted Answer

Step 1: From the given relations we have $\prod_{i=1}^{n} \cos a_i = \prod_{i=1}^{n} \sin a_i$

Step 2: $\prod_{i=1}^{n} \cos^2 a_i = \prod_{i=1}^{n} (\cos a_i \sin a_i) = \prod_{i=1}^{n} \frac{\sin 2a_i}{2}$

Step 3: Since $0 < a_i < \frac{\pi}{2}$, we have $0 < 2a_i < \pi$, so $\prod_{i=1}^{n} \cos^2 a_i < \frac{1}{2^n}$ as the maximum value of $\sin 2a_i$ is 1 for all $i$.

Step 4: Therefore $\prod_{i=1}^{n} \cos a_i < \frac{1}{2^{n/2}}$

∴ The maximum value is $\frac{1}{2^{n/2}}$

Question:

Step-by-Step Solution

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