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$\cos A \cos 2A \cos 4A....... \cos 2^{n-1} A=$

\[\begin {array} {1 1} (1)\;\frac{sin 2^n A}{2^n \sin A} & \quad (2)\;\frac{2^n \sin 2^n A}{\sin A} \\ (3)\;\frac{2^n \sin A}{sin 2^n A} & \quad (4)\;\frac{\sin A}{2^n \sin 2^n A} \end {array}\]
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$ (1)\;\frac{sin 2^n A}{2^n \sin A}$
answered Nov 7, 2013 by pady_1
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