﻿ Prove that: cos(π/15) cos(2π/15) cos(3π/15) cos(4π/15) cos(5π/15) cos(6π/5) cos(7π/15) = 1/128 \cos\dfrac{\pi}{15} \cos\dfrac{2\pi}{15} \cos\dfrac{3\pi}{15} \cos\dfrac{4\pi}{15}

Question

Prove that: cos(π/15) cos(2π/15) cos(3π/15) cos(4π/15) cos(5π/15) cos(6π/5) cos(7π/15) = 1/128

\cos\dfrac{\pi}{15} \cos\dfrac{2\pi}{15} \cos\dfrac{3\pi}{15} \cos\dfrac{4\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} \cos\dfrac{7\pi}{15}=\dfrac{1}{128}

Collected in the board: Trigonometry

Steven Zheng posted 8 hours ago

Repeat using of double angle identity for sines function

\cos\dfrac{\pi}{15} \cos\dfrac{2\pi}{15} \cos\dfrac{3\pi}{15} \cos\dfrac{4\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} \cos\dfrac{7\pi}{15}

=\dfrac{\Big( 2\sin\dfrac{\pi}{15}\cos\dfrac{\pi}{15}\Big) \cos\dfrac{2\pi}{15} \cos\dfrac{3\pi}{15} \cos\dfrac{4\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} \cos\dfrac{7\pi}{15} }{2\sin\dfrac{\pi}{15}}

=\dfrac{\sin\dfrac{2\pi}{15}\cos\dfrac{2\pi}{15} \cos\dfrac{3\pi}{15} \cos\dfrac{4\pi}{15} \cos\dfrac{15\pi}{15} \cos\dfrac{6\pi}{15} \cos\dfrac{7\pi}{15} }{2\sin\dfrac{\pi}{15}}

=\dfrac{\Big( 2\sin\dfrac{2\pi}{15}\cos\dfrac{2\pi}{15}\Big) \cos\dfrac{3\pi}{15} \cos\dfrac{4\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} \cos\dfrac{7\pi}{15} }{4\sin\dfrac{\pi}{15}}

==\dfrac{ \sin\dfrac{4\pi}{15} \cos\dfrac{3\pi}{15} \cos\dfrac{4\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} \cos\dfrac{7\pi}{15} }{4\sin\dfrac{\pi}{15}}

=\dfrac{\Big( 2 \sin\dfrac{4\pi}{15}\cos\dfrac{4\pi}{15} \Big) \cos\dfrac{3\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} \cos\dfrac{7\pi}{15} }{8\sin\dfrac{\pi}{15}}

=\dfrac{\sin\dfrac{8\pi}{15} \cos\dfrac{3\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} \cos\dfrac{7\pi}{15} }{8\sin\dfrac{\pi}{15}}

=\dfrac{\Big( 2\sin\dfrac{8\pi}{15} \cos\dfrac{7\pi}{15}\Big) \cos\dfrac{3\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} }{16\sin\dfrac{\pi}{15}}

=\dfrac{ \Big[ \cos\Big( \dfrac{8\pi}{15}- \dfrac{7\pi}{15} \Big) - \cos\Big(\dfrac{8\pi}{15} + \dfrac{7\pi}{15}\Big) \Big] \cos\dfrac{3\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} }{16\sin\dfrac{\pi}{15}}

=\dfrac{ \Big( \sin\dfrac{\pi}{15} - \sin \pi \Big) \cos\dfrac{3\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} }{16\sin\dfrac{\pi}{15}}

=\dfrac{ \cancel{\sin\dfrac{\pi}{15}} \cos\dfrac{3\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} }{16\cancel{\sin\dfrac{\pi}{15}}}

=\dfrac{ \Big( 2\sin\dfrac{3\pi}{15} \cos\dfrac{3\pi}{15}\Big) \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} }{32 \sin\dfrac{3\pi}{15}}

=\dfrac{ \sin\dfrac{6\pi}{15} \cos\dfrac{5\pi}{15} \cos\dfrac{6\pi}{15} }{32 \sin\dfrac{3\pi}{15}}

=\dfrac{ \Big( 2\sin\dfrac{6\pi}{15}\cos\dfrac{6\pi}{15} ) \cos\dfrac{5\pi}{15} }{64 \sin\dfrac{3\pi}{15}}

=\dfrac{ \sin\dfrac{12\pi}{15} \cos\dfrac{5\pi}{15} }{64 \sin\dfrac{3\pi}{15}}

=\dfrac{\sin\Big( \pi-\dfrac{3\pi}{15} \Big) \cos\dfrac{5\pi}{15} \\ }{64 \sin\dfrac{3\pi}{15}}

==\dfrac{\cancel{\sin\ \dfrac{3\pi}{15} } \cos\dfrac{5\pi}{15} \\ }{64\cancel{ \sin\dfrac{3\pi}{15}}}

=\dfrac{1}{64} \cos\dfrac{\pi}{3} =\dfrac{1}{128}

Steven Zheng posted 8 hours ago

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