What is the maximum value of sinA*sinB*sinC*sinD when tanA*tanB*tanC*tanD = 1?

Question

{A,B,C,D} belong to (0,pi/2)

Answer 1

As Ken M suggested, the maximum is 1/4.

Proving this is infuriatingly difficult to do. Ken M did not prove it. The brute force way to do it is to determine the value of D, given that Tan(A)Tan(B)Tan(C)Tan(D) = 1. Having done that, then we have a function F(A,B,C) to find the maximum:

F(A,B,C) = Sin(A)Sin(B)Sin(C)*
(ArcTan(1/Tan(A)Tan(B)Tan(C)))

Then we find the partial derivatives for each A, B, C, all of them set to 0, so that we have a system of 3 equations to solve for A, B, C. Space doesn't permit the actual work, but solution is A = B = C = π/4, and so the maximum really is 1/4.

It turns out that the function F(A,B,C) has a far from trivial structure, particularly where either A, B, or C is close to (nπ)/2, where n is any integer.

If anyone can offer a simpler way to prove this, I would really love to see it.

Answer 2

Since tan x = sin x / cos x, the condition that tanA*tanB*tanC*tanD = 1is equivalent to the condition that sin A * sin B * sin C * sin D = cos A * cos B * cos C * cos D. If sine A = cos A then A = pi / 4. Indeed tan pi/4 = 1. So it would seem that the maximum value of sinA*sinB*sinC*sinD is obtained when all the angles equal pi/4, giving a product of 1/4.

Answer 3

The maximum value of sin can be 1 and the minimum value can be -1. Since all sin are in multiply so the maximum value of your problem can be only 1