Sines and pies

Geometry Level 5

sin π 14 sin 3 π 14 sin 5 π 14 sin 7 π 14 sin 9 π 14 sin 11 π 14 sin 13 π 14 = 2 n \large \sin\dfrac{\pi}{14}\sin\dfrac{3{\pi}}{14}\sin\dfrac{5{\pi}}{14}\sin\dfrac{7{\pi}}{14}\sin\dfrac{9{\pi}}{14}\sin\dfrac{11{\pi}}{14}\sin\dfrac{13{\pi}}{14}\ = 2^n

Find n n .


The answer is -6.

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1 solution

Chew-Seong Cheong
May 16, 2016

sin π 14 sin 3 π 14 sin 5 π 14 sin 7 π 14 sin 9 π 14 sin 11 π 14 sin 13 π 14 As sin x = cos ( π 2 x ) and cos ( x ) = cos x = cos 6 π 14 cos 4 π 14 cos 2 π 14 cos 0 cos 2 π 14 cos 4 π 14 cos 6 π 14 = ( cos π 7 cos 2 π 7 cos 3 π 7 ) 2 As cos x = cos ( π x ) = ( cos π 7 cos 2 π 7 cos 4 π 7 ) 2 = ( sin π 7 cos π 7 cos 2 π 7 cos 4 π 7 sin π 7 ) 2 = ( sin 2 π 7 cos 2 π 7 cos 4 π 7 2 sin π 7 ) 2 = ( sin 4 π 7 cos 4 π 7 4 sin π 7 ) 2 = ( sin 8 π 7 8 sin π 7 ) 2 As cos x = cos ( π x ) = ( sin π 7 8 sin π 7 ) 2 = ( 1 8 ) 2 = 1 2 6 \sin\dfrac{\pi}{14} \sin\dfrac{3{\pi}}{14} \sin\dfrac{5{\pi}}{14} \sin\dfrac{7{\pi}}{14} \sin\dfrac{9{\pi}}{14} \sin\dfrac{11{\pi}}{14} \sin\dfrac{13{\pi}}{14} \quad \quad \small \color{#3D99F6}{\text{As } \sin x = \cos \left(\dfrac{\pi}{2}-x\right) \text{ and } \cos (-x) = \cos x} \\ = \cos \dfrac{6\pi}{14}\cos \dfrac{4\pi}{14}\cos \dfrac{2\pi}{14}\cos 0 \cos \dfrac{2\pi}{14}\cos \dfrac{4\pi}{14}\cos \dfrac{6\pi}{14} \\ = \left(\cos \dfrac{\pi}{7}\cos \dfrac{2\pi}{7}\color{#D61F06}{\cos \dfrac{3\pi}{7}} \right)^2 \quad \quad \small \color{#D61F06}{\text{As } \cos x = - \cos (\pi - x)} \\ = \left(\color{#D61F06}{-}\cos \dfrac{\pi}{7}\cos \dfrac{2\pi}{7} \color{#D61F06}{\cos \dfrac{4\pi}{7}} \right)^2 \\ = \left(- \dfrac{\sin \frac{\pi}{7} \cos \frac{\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}}{\sin \frac{\pi}{7}} \right)^2 \\ = \left(- \dfrac{\sin \frac{2\pi}{7} \cos \frac{2\pi}{7} \cos \frac{4\pi}{7}}{2 \sin \frac{\pi}{7}} \right)^2 \\ = \left(- \dfrac{\sin \frac{4\pi}{7} \cos \frac{4\pi}{7}}{4 \sin \frac{\pi}{7}} \right)^2 \\ = \left(- \dfrac{\color{#D61F06}{\sin \frac{8\pi}{7}}}{8 \sin \frac{\pi}{7}} \right)^2 \quad \quad \small \color{#D61F06}{\text{As } \cos x = - \cos (\pi - x)} \\ = \left(\dfrac{\color{#D61F06}{\sin \frac{\pi}{7}}}{8 \sin \frac{\pi}{7}} \right)^2 \\ = \left(\dfrac{1}{8} \right)^2 \\ = \frac{1}{2^6}

n = 6 \implies n = \boxed{-6}

How sin x =cos(π\2-x)

Praful Jain - 5 years ago

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Please read up this link . Also read up our wiki on Trigonometry .

Chew-Seong Cheong - 5 years ago

Very nice approach

Sanchit Sharma - 5 years, 1 month ago

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