Clarification/Follow-up by Jim.McGinness on 10/14/06 8:05 am:
Too hard for me. I only got so far as to realize that the last term contains angles that are (90 - n) degrees from the angles in the first three terms, so
with A=11
sin^2(A) + sin^2(2A) + sin^2(3A) + 2*cos(A)*cos(2A)*cos(3A)
is to be shown to be necessarily equal to 2. I was left wondering if this is independent of what A is, so I tried it with A=2 and also got the answer 2. That tells me we're engaged with one of the myriad trig identities, but my table of multiple angle and power identities does not give me a clue about the path to be followed to prove this particular identity.
Clarification/Follow-up by ontadian on 10/14/06 11:26 am:
Hi Jim,
Yes it is a tough one, and does require trig. identities for its solution. I want to leave it as posted for now, but if you would like the solution sooner, send me an e-mail (to ontadian@hotmail.com)
Regards
Bill
Clarification/Follow-up by Jim.McGinness on 10/14/06 11:56 pm:
This was keeping me awake....
After the above, I did the substitutions
sin x = (1/2i)(e^ix - e^-ix)
and
cos x = (1/2)(e^ix + e^-ix)
and, multiplying things out, saw everything cancel out except for the places where e^iy had been multiplied by e^-iy, giving a 1. The coefficients of all those ŕ"s added up to exactly 2.
Aside from being a brain-teaser, does this identity have some real-world uses?
Clarification/Follow-up by Jim.McGinness on 10/15/06 12:51 am:
Ooops, I see that Bluto did something very like this a couple of years ago.
