wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> prove int
        
(Message started by: tony123 on Sep 22nd, 2007, 1:15pm)
Title: prove int
Post by tony123 on Sep 22nd, 2007, 1:15pm
cos (-89)+cos (-87) +.......+cos87 +cos 89 =csc 1


tanČ(pi/7) +tanČ(2pi/7) +tanČ(3pi/7) =4
Title: Re: prove int
Post by Michael_Dagg on Sep 22nd, 2007, 1:59pm
Anyone know what "prove int" means? -- I don't.
Title: Re: prove int
Post by Sameer on Sep 22nd, 2007, 2:03pm

on 09/22/07 at 13:59:24, Michael_Dagg wrote:
Anyone know what "prove int" means? -- I don't.


tony123 has a bad habit of not providing good titles. Essentially his problems are prove LHS = RHS and in this case two problems...
Title: Re: prove int
Post by ThudanBlunder on Sep 22nd, 2007, 6:51pm

on 09/22/07 at 13:59:24, Michael_Dagg wrote:
Anyone know what "prove int" means? -- I don't.

I would guess he means Prove It.
Title: Re: prove int
Post by srn347 on Sep 22nd, 2007, 10:01pm
Or maybe he meant prove intelligently.
Title: Re: prove int
Post by tony123 on Sep 23rd, 2007, 1:43am
sory all  :-*
i mean prove  LHS = RHS  in  two problems  :-[
Title: Re: prove int
Post by Barukh on Sep 24th, 2007, 9:38am
The first is easy. The second looks much more difficult (and interesting).

BTW, shouldn't it be 21?
Title: Re: prove int
Post by Aryabhatta on Sep 24th, 2007, 1:14pm
For the second one:

[hide]
If a = pi/7

then
tan a, tan 2a, ..., tan 6a are roots of the polynomial

P(x) = x6 - 21x4 + 35x2 - 7

Sum of squares of roots of P(x) is 2*21.

Hence the required value is half of that = 21.
[/hide]
Title: Re: prove int
Post by Barukh on Sep 25th, 2007, 9:56am

on 09/24/07 at 13:14:46, Aryabhatta wrote:
For the second one:

[hide]
If a = pi/7

then
tan a, tan 2a, ..., tan 6a are roots of the polynomial

P(x) = x6 - 21x4 + 35x2 - 7[/hide]

Why? That's not obvious!
Title: Re: prove int
Post by pex on Sep 25th, 2007, 10:34am

on 09/25/07 at 09:56:34, Barukh wrote:
Why? That's not obvious!
Yes, I do think it is. Repeatedly using the addition formula for tangents, we have

tan(7t) = tan t * [tan6t - 21tan4t + 35tan2t - 7] / [7tan6t - 35tan4t + 21tan2t - 1].

Now obviously, for t = pi/7, 2pi/7, ..., 7pi/7, we have tan(7t) = 0 and thus, these are the roots of

tan t * [tan6t - 21tan4t + 35tan2t - 7] = 0.

Put differently, tan(pi/7), tan(2pi/7), ..., tan(7pi/7) are the roots of

x * [x6 - 21x4 + 35x2 - 7] = 0.

Since tan(7pi/7) = 0, we may divide by x to leave an equation of which tan(pi/7), tan(2pi/7), ..., tan(6pi/7) are the roots:

x6 - 21x4 + 35x2 - 7 = 0.
Title: Re: prove int
Post by Aryabhatta on Sep 25th, 2007, 11:32am

on 09/25/07 at 09:56:34, Barukh wrote:
Why? That's not obvious!


It is not, but I didn't want to spoil it for others. pex has explained a way to get it. I did something slightly different, (but essentially same): [hide] tan 3a + tan 4a = 0 [/hide]
Title: Re: prove int
Post by Eigenray on Sep 25th, 2007, 2:43pm
And here is how I did it: [hide]sin(7t)/[sin(t)cos6(t)] = 0[/hide].

Actually, at first I did [hide]sin(7t)/sin(t)=0, got a polynomial in x=cos2t, and solved for http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif 1/cos2(khttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/7) = 24[/hide].

Actually, before that I expressed it in terms of http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif = ehttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/suppi.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/supi.gif/7, and "saw" that it was divisible by http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif14(http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/zeta.gif)=0.  But that is not a good solution.
Title: Re: prove int
Post by Barukh on Sep 25th, 2007, 11:37pm
Nice!

The result may be generalized for any odd number 2n+1.

What about the first one?
Title: Re: prove int
Post by tony123 on Sep 26th, 2007, 3:43am

on 09/22/07 at 13:15:40, tony123 wrote:
cos (-89)+cos (-87) +.......+cos87 +cos 89 =csc 1


tanČ(pi/7) +tanČ(2pi/7) +tanČ(3pi/7) =21
Title: Re: prove int
Post by Aryabhatta on Sep 26th, 2007, 11:45am

on 09/25/07 at 23:37:41, Barukh wrote:
What about the first one?


Isn't the "arithmetic series" summation of cosines well known?

cos 0 + cos h + cos 2h + ... + cos nh = (sin((n+1)h/2) * sin (nh/2))/sin(h/2)

(can be done easily using complex numbers I think)
Title: Re: prove int
Post by Sameer on Sep 26th, 2007, 12:59pm

on 09/26/07 at 11:45:05, Aryabhatta wrote:
Isn't the "arithmetic series" summation of cosines well known?

cos 0 + cos h + cos 2h + ... + cos nh = (sin((n+1)h/2) * sin (nh/2))/sin(h/2)

(can be done easily using complex numbers I think)


Correct. You can use the Cis summation method nicely demonstrated by iyerkri

http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_putnam;action=display;num=1189578447


Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board