Complex triangle equality & Sin(ntheta)/Sin(theta)

[tiny-tim]

Looking at RHS, the first term just disappears. The second term though, again, can I say much about it without knowing n?

try it first (ie θ = π/2) for n odd, then the RHS is … ?

then try it for n even, but first divide both sides by cosθ

 

[Verdict]

Alright, so I tried for n = 5, which gives 1/16 on RHS and 1 on LHS, so indeed the factor works out as 2^(n-1). Not a general proof of course.. Still have to figure that one out. Is there an identity for this?
But for the even, how can you divide by cos theta if you set it is equal to 0? (pi/2 for theta)

 

[tiny-tim]

Alright, so I tried for n = 5, which gives 1/16 on RHS and 1 on LHS, so indeed the factor works out as 2^(n-1). Not a general proof of course.. Still have to figure that one out. Is there an identity for this?

hmm … even for odd n, it's not as easy as i thought it would be

But for the even, how can you divide by cos theta if you set it is equal to 0? (pi/2 for theta)

you can divide both sides by cosθ first, to give you a general equation, and then put cosθ = 0

(and I'm off to bed now :zzz:)

 

[Verdict]

Yeah, I can't really see an easy method too :(
And yes, dividing by cosine first is the only thing that makes sense I suppose, but that makes it a bit more complicated than it was! It was all so pretty..

Anyway, thanks a lot for the help, I think I will give it a rest for now too and look at it again tomorrow.

 

[Verdict]

The professor suggested looking at Chebyshev polynomials, does that make any sense to you?