Trigonometric product

Here n=1 to n = 7(upper limit) not 15

The 15 th roots of unity are given by

e^{(\frac{2k\pi }{15} )}

where k = -7,-6,-5,...0,1,...5,6,7

so,\frac{z^{15}-1}{z-1} =
(z-cos\frac{2\pi }{15}- i sin \frac{2\pi }{15})(z-cos\frac{2\pi }{15}+ sin\frac{2\pi }{15})
(z-cos\frac{4\pi }{15}- i sin \frac{4\pi }{15})(z-cos\frac{4\pi }{15}+ sin\frac{4\pi }{15})
.....
(z-cos\frac{14\pi }{15}- i sin \frac{14\pi }{15})(z-cos\frac{14\pi }{15}+ sin\frac{14\pi }{15})

=(z^{2}- 2cos\frac{2\pi }{15}+1 )(z^{2}- 2cos\frac{4\pi }{15}+1 )(z^{2}- 2cos\frac{6\pi }{15}+1 )...(z^{2}- 2cos\frac{14\pi }{15}+1 )

Now,putting z=-1 we get

1= 4^{7} \prod_{n=1}^{7}{}cos^{2}\frac{n\pi }{15}

(using the identity cos2x = 2cos^{2}x -1)

so \prod_{n=1}^{7}{}cos\frac{n\pi }{15} = \frac{1}{2^{7}}

(-\frac{1}{2^{7}} is not taken because all terms in the product are individually positive)

exact!