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        riddles >> easy >> Square and Sum the Odd Tans
        
(Message started by: TenaliRaman on Dec 3^rd, 2003, 11:33am)

Title: Square and Sum the Odd Tans
Post by TenaliRaman on Dec 3^rd, 2003, 11:33am

Prove
tan²(1^o) + tan²(3^o) + tan²(5^o) + .... + tan²(89^o)
is an integer.

Title: Re: Square and Sum the Odd Tans
Post by Eigenray on Dec 3^rd, 2003, 1:51pm

You could [hide]explicitly construct a polynomial whose roots are the 45 terms in the sum[/hide], but I don't know if I'd call that "easy," so maybe there's a simpler way.

Title: Re: Square and Sum the Odd Tans
Post by TenaliRaman on Dec 3^rd, 2003, 10:11pm

you have the right approach and there is an easier way to go about. :)

Hint: (quite vague)[hide]ask De Moivre for help[/hide]

Title: Re: Square and Sum the Odd Tans
Post by Eigenray on Dec 4^th, 2003, 11:55pm

If t = (2n+1)^o, then[hide]
0 = cos(90t) = Re[ e^i90t ] = Re[ (cos t + isin t)⁹⁰ ] = [sum]_k=0⁴⁵ (⁹⁰C_2k)(-1)^kcos^90-2kt sin^2kt.
Dividing by cos⁹⁰t shows that x = tan²t is a root of the polynomial
f(x) = [sum]_k=0⁴⁵ (⁹⁰C_2k)(-1)^kx^k = 0.
Since f can have at most 45 roots, its roots are exactly {tan²(2n+1)^o, n=0..44}, which therefore sum to ⁹⁰C₂ = 4005[/hide], an integer.

Title: Re: Square and Sum the Odd Tans
Post by TenaliRaman on Dec 5^th, 2003, 5:36am

Awesome!!! 8)

Title: Re: Square and Sum the Odd Tans
Post by Barukh on Dec 5^th, 2003, 6:35am

I agree with TenaliRaman, it's absolutely brilliant! :D

Eigenray, could you please write a bit about how did you arrive at this solution?

TenaliRaman, do you think this problem belongs to easy section?

Title: Re: Square and Sum the Odd Tans
Post by TenaliRaman on Dec 5^th, 2003, 11:31am

I think it does belong to easy section, doesn't it?
i mean if i could solve it in one and a half days, it should be easy.