Maximum value trig problem

[ScholMaths]

Can someone please help me with this question from a very very old exam paper:

Prove that sin^2((A+B)/2) - sin^2((A-B)/2) = sinAsinB

and hence show that if A,B and C are three variable acute angles such that A+B+C is a constant then the greatest value of the product of sinAsinBsinC occurs when A=B=C.

I can prove the first part by using the result sin^2(A) = 1/2 (1-cos2A) and then solving through.

How is the first part then used to solve the second part of the question?

 

[Deleted member 4993]

Can someone please help me with this question from a very very old exam paper:

Prove that sin^2((A+B)/2) - sin^2((A-B)/2) = sinAsinB

and hence show that if A,B and C are three variable acute angles such that A+B+C is a constant then the greatest value of the product of sinAsinBsinC occurs when A=B=C.

I can prove the first part by using the result sin^2(A) = 1/2 (1-cos2A) and then solving through.

How is the first part then used to solve the second part of the question?

Have you taken a course in Calculus?

For the first part, easier method would be to:

sin^2((A+B)/2) - sin^2((A-B)/2)

= [sin((A+B)/2) - sin((A-B)/2)] * [sin((A+B)/2) + sin((A-B)/2)]

= [2*sin(B/2)*cos(A/2)]*[2*sin(A/2)*cos(B/2]

= sin(2*A/2)*sin(2*B/2)

= sin(A) * sin(B)

 

[ScholMaths]

Yes - I have done a course in calculus, although this question came from the non-calculus paper "a" (trig + algebra).
So, I guess the next step is to use calculus to maximise:

sin(k-A-B)*[sin^2((A+B)/2) - sin^2((A-B)/2)], as A+B+C =constant, k ?

Any help from here please?

 

[Novice]

I cannot completely satisfy you due to my limited knowledge, but I'll try.
For sinAsinB to be maximum, sin^2((A+B)/2) - sin^2((A-B)/2) should be maximum, which is true when the negative term, sin^2((A-B)/2) is minimum, i.e, it is equal to zero (it cannot be negative as sin^2((A-B)/2 is a square). This implies that A-B/2=0, or A-B=0, or A=B. Similarly we can prove that maximum values of sinBsinC and sinCsinA are achieved when B=C and C=A respectively, thus implying, that A=B=C.

Explanation: For the square root of the product of sinAsinB, sinBsinC, and sinCsinA (i.e. (sinAsinBsinC)^2 to be maximum, each of sinAsinB, sinBsinC, and sinCsinA should be maximum, which we have shown is achieved when A=B=C.
THEOREM: If a+b+c are given to be constant, any cyclic function in a, b, and c attains its minimum OR maximum value at a=b=c( for convenience I have used three variables, but this property is true for any number of variables). Why, I do not know.