(\frac12+cos\frac{\pi}{20})(\frac12+cos\frac{3\pi}{20})(\frac12+cos\frac{9\pi}{20})(\frac12+cos\frac{27\pi}{20})=\frac1{16}
This is another problem again from Titu Andreescu's book...
(Hint: Chapter: Telescopic products and Products)
62Try this too.. not very difficult if you see the hints in the question statement itself..
1\frac{(1+2cos\theta )(1+2cos3\theta )(1+2cos9\theta )(1+2cos27\theta )}{16}
now
(1+2cos\theta )=1+z+\frac{1}{z}
(1+2cos\theta )=(1+z+z^{2})/z
\frac{(1-z)(1+z+z^{2})(1+z^{3}+z^{6})(1+z^{9}+z^{18})(1+z^{27}+z^{54})}{(1-z)z^{40}}
numerator collapses.........(thanks to prophet sir's tip. http://targetiit.com/iit-jee-forum/posts/trigonometry-11490.html)
z81=z
\frac{(1-z)}{(1-z)z^{40}}
=1/16
62Good work.. that hint was because there is an alternate way of doing this one too :)
if you can think of cos 81/20 pi (and from where it could feature!)