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Let $ P =\prod^{7}_{k = 1}\cos\left(\frac{k\pi}{15}\right)$ and $ Q =\prod^{7}_{k = 1}\sin\left(\frac{k\pi}{15}\right)$ obviously $ Q\neq0$, by $ \sin\frac{k\pi}{15}\cos\frac{k\pi}{15}=\frac{1}{2}\sin\frac{2k\pi}{15}$ we have: $ PQ =\frac{1}{2^{7}}\prod^{7}_{k = 1}\sin\left(\frac{2k\pi}{15}\right)$ but $ \sin\frac{8\pi}{15}=\sin\frac{15\pi-7\pi}{15}=\sin\left(\pi-\frac{7\pi}{15}\right) =\sin\frac{7\pi}{15}$ $ \sin\frac{10\pi}{15}=\sin\frac{5\pi}{15}\;\;,\sin\frac{12\pi}{15}=\sin\frac{3\pi}{15}\;\;,\sin\frac{14\pi}{15}=\sin\frac{\pi}{15}$ As a consequence: $ \prod^{7}_{k = 1}\sin\left(\frac{2k\pi}{15}\right) =\prod^{7}_{m = 1}\sin\left(\frac{m\pi}{15}\right) = Q$ and so $ PQ =\frac{1}{2^{7}}Q$ and $ \boxed{P = 2^{-7}}$.

Solution. Let $ z = e^{i\frac{2\pi}{15}}$ be the 15-th root of unity. Clearly, $ \cos\frac{k\pi}{15}=\frac{z^{\frac{k}{2}}+\overline{z}^{\frac{k}{2}}}{2}=\frac{z^{k}+1}{2z^{\frac{k}{2}}}$. Hence, \[ P=\prod_{k=1}^{7}\cos\frac{k\pi}{15}=\frac{1}{2^{7}z^{14}}\prod_{k=1}^{7}(z^{k}+1).\] Using $ \sum_{k = 0}^{14}z^{k}= 0$, we obtain that $ P =\frac{1}{128}$. $ \Box$

Uhhh...it does not follow that immediately. $ \sum_{0}^{14}x^{k}\not|\prod_{1}^{7}x^{k}+1$ You need a stronger factorization of $ x^{15}= 1$. Since $ 15$ is not prime, not all roots are primitive.