1can u plz solve it......im not getting...
if....\theta =\frac{2\pi }{2009}.... then.....cos\theta cos 2\theta cos 3\theta ..............cos 1004\theta ........is equal to .....
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7 Answers
23aniruddha i think u made a mistake ..the method u r applying is true when the angles inside cosa are in GP with cr =2
11So I find none interested in the method I mentioned....gr8!
\frac{x^{2n+1}-1}{x-1}=\left(x^2-2xcos\frac{2\pi}{2n+1}+1 \right)\left(x^2-2xcos\frac{4\pi}{2n+1}+1\right)....\left(x^2-2xcos\frac{2n\pi}{2n+1}+1 \right) As x →-1, we get 1=4^n\prod_{k=1}^{n}{cos^2 \frac{k\pi}{2n+1}}
Which finally gives \prod_{k=1}^{n}{cos\frac{k\pi}{2n+1}}=\frac{1}{2^n}
Remember this!
So now it becomes cake-walk......
If I'm not wrong in my calculations, ans = \frac{1}{2^{1004}}.....
Job done!
1someone please explain soumik bhaiya's first line of the solution.
thanks nishant bhaiya.
62The first line is basically the observation that the roots of x2n+1-1 =0 are of the form eiθ
after that he has clubbed together the conjugates...
(x-a)(x-a) = (x2 - (a+a)x+aa)
and x-1 is one factor taken to the bottom (divided on both sides)