Question

If \[ \prod_{i=1}^{n} \tan (\alpha_i) = 1 \forall \alpha_i \in \left[0, \frac{\pi}{2}\right] \text{where} i = 1, 2, 3, \dots, n. \] Then the maximum value of \[ \prod_{i=1}^{n} \sin (\alpha_i) \] is

• A. \( \frac{1}{2^n} \)
• B. \( \frac{1}{2^{n/2}} \)
• C. 1
• D. None of these

Hint:

For maximizing or minimizing product expressions involving trigonometric functions, it's often helpful to set all angles equal when symmetry exists.

Answer 1

The Correct Option is B

We are given that: \[ \prod_{i=1}^{n} \tan (\alpha_i) = 1 \text{for} \alpha_i \in \left[0, \frac{\pi}{2}\right]. \] The tangent of an angle \( \alpha \) is given by: \[ \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} \] Thus, the given condition becomes: \[ \prod_{i=1}^{n} \frac{\sin (\alpha_i)}{\cos (\alpha_i)} = 1 \] This simplifies to: \[ \frac{\prod_{i=1}^{n} \sin (\alpha_i)}{\prod_{i=1}^{n} \cos (\alpha_i)} = 1 \] Therefore: \[ \prod_{i=1}^{n} \sin (\alpha_i) = \prod_{i=1}^{n} \cos (\alpha_i) \] Now, for the maximum value of \( \prod_{i=1}^{n} \sin (\alpha_i) \), the optimal condition occurs when all the angles \( \alpha_i \) are equal. Let \( \alpha_1 = \alpha_2 = \dots = \alpha_n = \alpha \). Then the condition becomes: \[ \tan(\alpha)^n = 1 \Rightarrow \tan(\alpha) = 1 \Rightarrow \alpha = \frac{\pi}{4} \] Substituting \( \alpha = \frac{\pi}{4} \) into the expression for \( \sin(\alpha) \): \[ \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \] Thus: \[ \prod_{i=1}^{n} \sin (\alpha_i) = \left( \frac{\sqrt{2}}{2} \right)^n = \frac{1}{2^{n/2}} \] Therefore, the maximum value of \( \prod_{i=1}^{n} \sin (\alpha_i) \) is \( \boxed{\frac{1}{2^{n/2}}} \).