Polynomial products

Question

This problem $$ \large \displaystyle\prod \limits^{14}_{k=1}\cos \left( \frac{k \pi }{15} \right) =\ ? $$ I solved it in this way $$ x = \displaystyle \prod \limits^{14}_{k=1}\cos \left( \frac{k\pi }{15} \right) y \\ = \displaystyle\prod \limits^{14}_{k=1}\sin\left( \frac{k\pi }{15} \right) x. y\\ =\displaystyle\prod \limits^{14}_{k=1}\sin\left( \frac{k\pi }{15} \right) \cos \left( \frac{k\pi }{15} \right) x. y \\ = \displaystyle\prod_{k=1}^{14} \frac{1}{2} \sin\left( \frac{2k\pi}{15} \right) \\ = \bigg(\frac{1}{2}\bigg)^{14}\sin \left( \frac{2\pi }{15} \right) \sin \left( \frac{4\pi }{15} \right) \cdots \sin \left( \frac{14\pi }{15} \right) \sin \left( \frac{16\pi }{15} \right) \sin \left( \frac{18\pi }{15} \right) \cdots \sin \left( \frac{28\pi }{15} \right) \\ \text {Now} :\sin \left( \frac{16\pi }{15} \right) = \left( -1\right) \sin \left( \frac{\pi }{15} \right) \& \sin \left( \frac{18\pi }{15} \right) \\ = \left( -1\right) \sin \left( \frac{3\pi }{15} \right)\& \cdots \&\sin \left( \frac{28\pi }{15} \right)\\ = \left( -1\right) \sin \left( \frac{13\pi }{15} \right) x. y \\ = \left( \frac{1}{2} \right) ^{14}\left( -1\right) ^{7} \bigg[ \sin \left( \frac{2\pi }{15} \right) \sin \left( \frac{4\pi }{15} \right) \cdots \sin \left( \frac{14\pi }{15} \right) \sin \left( \frac{\pi }{15} \right) \sin \left( \frac{3\pi }{15} \right) \cdots \sin \left( \frac{13\pi }{15} \right) \bigg] x. y\\ =\left( \frac{1}{2} \right) ^{14}\left( -1\right) ^{7}. y \text{ so} : x =-\left( \frac{1}{2} \right) ^{14} $$ the question is how I can solve it by Chebyshev Polynomials??? . any help will appreciate