A difficult trigonometric identity for genius?

Question

If a+b+c=180 (degrees)
Prove that
sin^3 a+sin^3 b+ sin^3 c=3cos(a/2)cos(b/2)cos(c/2)+ cos(3a/2)cos(3b/2)cos(3c/2)

Please show me each step of your working.

Thank you very much in advance.

P.S.this is a really brain-storming problem ;-)

It would be thankful for me to know how to prove the left hand side is equal to right hand side,with useful steps.

Answer 1

This is a hell of a question. In your reply please tell us what grade this is for as I can't imagine giving this to any of my students.

Firstly we can write c = 180 - a - b

So the question becomes:
sin³a+sin³b+ sin³(180-a-b) = 3cos(a/2)cos(b/2)cos((180-a-b)/2)+ cos(3a/2)cos(3b/2)cos(3(180-a-b)/2)
Or simplifying:
sin³a+sin³b+ sin³(a+b) = 3cos(a/2)cos(b/2)sin((a+b)/2) - cos(3a/2)cos(3b/2)sin(3(a+b)/2)
As cos(90-x) = sinx and cos(270-x) = - sinx due to the periodicity of cos and sin.

There is heap of rules/formulae you need to know/use in this.
I suggest you look them up if you don't know them.

Products to sums:
cosAcosB = 1/2(cos(A+B) + cos(A-B)
cosAsinB = 1/2(sin(A+B) - sin(A-B))
sinBcosA = 1/2(sinA+B) + sin(A+B))
sinAsinB = 1/2(cos(A-B) + cos(A+B))

A special cases is: sinAcosA = 1/2sin2A

Sums to products:
sinC + sinD = 2sin((C+D)/2)cos((C-D)/2)
sinC - sinD = 2cos((C+D)/2)sin((C-D)/2)
cosC + cosD = 2cos((C+D)/2)cos((C-D)/2)
cosC - cosD = 2sin((C+D)/2)sin((D-C)/2)

Triple angle:
And lastly we will also need sin3A = 3sinA - 4sin³A


So looking at the right hand side of this:
3cos(a/2)cos(b/2)sin((a+b)/2) - cos(3a/2)cos(3b/2)sin(3(a+b)/2)

=3sin((a+b)/2) [ 1/2( cos(a/2+b/2) + cos(a/2-b/2) ) ] - sin(3(a+b)/2) [ 1/2 (cis(3a/2+3b/2) + cos(3a/2-3b/2) ) ]

=3/2sin((a+b)/2)cos((a+b)/2) + 3/2sin((a+b)/2)cos((a-b)/2) - 1/2 sin(3(a+b)/2)cos(3(a+b)/2) - 1/2sin(3(a+b)/2)cos(3(a-b)/2)

=3/4sin(a+b) + 3/4sina + 3/4sinb - 1/4sin(3(a+b)) - 1/4sin(3a) -1/4sin(3b)

=3/4sin(a+b) + 3/4sina + 3/4sinb - 1/4sin(3(a+b)) - 1/4sin(3a) -1/4sin(3b)

=3/4sin(a+b) + 3/4sina + 3/4sinb - 1/4[ 3sin(a+b) - 4sin³(a+b) ] - 1/4[ 3sina - 4sin³a] - 1/4[ 3sinb - 4sin³b ]

=3/4sin(a+b) + 3/4sina + 3/4sinb - 3/4sin(a+b) + sin³(a+b) - 3/4sina + sin³a - 3/4sinb + 4sin³b

=sin³a + sin³b + sin³(a+b)

Which is the left hand side as required.

Answer 2

Partial Solution: Usually it's better to work with less variables and it certainly is in this case. Isolating a in a+b+c = 180 yields a = 180 - (b+c). We can substitute this into the identity we are trying to prove to get

[sin(180 - (b+c))]^3 + (sinb)^3 + (sinc)^3 = 3cos(90 - (b+c)/2)cos(b/2)cos(c/2) + cos(270 - 3(b+c)/2)cos(3b/2)cos(3c/2). Although this looks uglier than what we had to begin with, it is easier to work with. We can simplify it some using the identies for sin(x-y) and cos(x-y) expressions: sin(x-y) = sinxcosy - sinycosx and cos(x-y) = cosxcosy + sinxsiny. Applying these we get

(sin(b+c))^3 + (sinb)^3 + (sinc)^3 = 3sin((b+c)/2)cos(b/2)cos(c/2) - sin(3(b+c)/2)cos (3b/2)cos(3c/2).

This is a little better but the two sides do not resemble each other yet. Note that we have products of sines and cosines on the right hand side. This strongly suggests that we use the trig product to sum formulas. Those are always tough to remember, so I'll provide a link with those below.

At this point you are going to have to get your hands dirty and use the product to sum formulae on the right hand side a couple times (until the cosines vanish). Also, when you derive it you may find the identity sin(3x) = 3sinx - 4(sinx)^3 useful. It simplifies nicely in the end (I checked).

Addition: See Ian's work for a complete solution. I intentionally offered a partial solution to give you the opportunity to see if you could finish from the eqn. I reduced it to and the tips below it. It looks intimidating, but a couple applications of the product to sum formulae transform the RHS into an equation with just sines.

Answer 3

Use sin(3x) = 3sin(x) - 4(sin(x))^3
Solve for (sin(x))^3 =(3/4)sin(x) - (1/4)sin(3x)
The problem now reduces to
(sin(A))^3+ (sin(B))^3 +(sin(C))^3
= (3/4)[sin(A)+sin(B)+sin(C)] - (1/4)[sin(3A)+sin(3B)+sin(3C)]
= (3/4)[{2sin((A+B)/2)cos((A-B)/2)} + {2sin(C/2)cos(C/2)}]
- (1/4)[{2sin(3(A+B)/2)cos(3(A-B)/2)} +{2sin(3C/2)cos(3C/2)}]
Given that A+B+C = 180_deg.
Hence (A+B)/2 = 90 - C/2
sin((A+B)/2)=sin(90 -C/2)=cos(C/2)
Also sin(3(A+B)/2) = -cos(3C/2)
Therefore: (sin(A))^3 +(sin(B))^3+ (sin(C))^3
= (3/4)[2cos(C/2)cos((A-B)/2) + 2cos(C/2)cos((A+B)/2)]
- (1/4)[-2cos(3(C/2)cos(3(A-B)/2) - 2cos(3(C/2)cos(3(A+B)/2)]
= (3/2)cos(C/2)[cos((A-B)/2)+cos((A+B)/2)]
+(1/2)cos(3C/2)[cos(3(A-B)/2 + cos(3(A+B)/2)]
= (3/2)cos(C/2)[2cos(A/2)cos(B/2)]
+(1/2)cos(3C/2)[2cos(3A/2)cos(3B/2)]
= 3cos(A/2)cos(B/2)cos(C/2)
+cos(3A/2)cos(3B/2)cos(3C/2)

Answer 4

Sinc Sind