[njana50]

[ebaines]

Hello njana. You probably would have gotten a quicker answer if you had posted this question either under homework help or in the mathematics section of the Science area. Anyway, here's one technique - although there may be other, more elegant ways to prove this:

1. Suppose the angle A is fixed at some constant. What values for B and C yield the greatest value for f=cos(A)cos(B)cos(C)? Since they must add to 180, you can express C as a function of B and A:

C = 180-A-B

So the product is:

f=cos(A)cos(B)cos(180-A-B)

Holding A constant, this function is maximized if B = 180-A-B, or B = (180-A)/2 (the proof is left to you, but essentially take the derivative of f with respect to B and set that =0).

2. Since C = 180-A-B, this means that C also equals (180-A)/2, and hence the max value of f occurs when B = C.

3. To find what value of A yields a maximum, find the max in the function:

f = cos(A)cos(B)cos(C) = cos(A) cos^2((180-A)/2)

Take the derivative of f with respect to A and set it to zero, and solve for A, and you'll find that A = 60. Consequently B = 60 and C = 60. Therefore f=cos(A)cos(B)cos(C) has a max value when all three angles = 60, and that value is 1/2 * 1/2 *1/2 = 1/8.

[voducdien]

You can not assume angle A is fixed. That's not allowed.

[ebaines]

You can not assume angle A is fixed. That's not allowed.

A is not assumed to be fixed. Go back and look at step 3. This is where the value of cos(A)cos(B) cos(C) is shown to be a max if A = 60 degrees, and that value is 1/8. Any other value for A yields a value of cos(A)cos(B)cos(C) that is smaller than that.