A geometry problem by Harry Jones

Geometry Level 4

sin 2 2 π 7 sin 2 π 7 + sin 2 4 π 7 sin 2 2 π 7 + sin 2 π 7 sin 2 4 π 7 = ? \large{\begin{aligned}\frac{\sin^2{\frac{2\pi }{7}}}{\sin^2{\frac{\pi }{7}}}+\frac{\sin^2{\frac{4\pi }{7}}}{\sin^2{\frac{2\pi }{7}}}+\frac{\sin^2{\frac{\pi }{7}}}{\sin^2{\frac{4\pi }{7}}}\end{aligned}} = \, ?


The answer is 5.

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1 solution

S = sin 2 2 π 7 sin 2 π 7 + sin 2 4 π 7 sin 2 2 π 7 + sin 2 π 7 sin 2 4 π 7 = sin 2 2 π 7 sin 2 π 7 + sin 2 4 π 7 sin 2 2 π 7 + ( sin 8 π 7 ) 2 sin 2 4 π 7 = n = 0 2 sin 2 2 n + 1 π 7 sin 2 2 n π 7 = n = 0 2 4 sin 2 2 n π 7 cos 2 2 n π 7 sin 2 2 n π 7 = n = 0 2 4 cos 2 2 n π 7 Note that: cos 2 x = 1 2 ( cos 2 x + 1 ) = 2 n = 0 2 ( cos 2 n + 1 π 7 + 1 ) = 2 ( cos 2 π 7 + cos 4 π 7 + cos 8 π 7 ) + 6 Note that: cos ( π x ) = cos x = 2 ( cos 5 π 7 + cos 3 π 7 + cos π 7 ) + 6 See note: k = 0 n 1 cos ( 2 k + 1 2 n + 1 π ) = 1 2 = 2 ( 1 2 ) + 6 = 5 \begin{aligned} S & = \frac{\sin^2{\frac{2\pi }{7}}}{\sin^2{\frac{\pi }{7}}}+\frac{\sin^2{\frac{4\pi }{7}}}{\sin^2{\frac{2\pi }{7}}}+\frac{\color{#3D99F6} \sin^2{\frac{\pi }{7}}}{\sin^2{\frac{4\pi }{7}}} \\ & = \frac{\sin^2{\frac{2\pi }{7}}}{\sin^2{\frac{\pi }{7}}}+\frac{\sin^2{\frac{4\pi }{7}}}{\sin^2{\frac{2\pi }{7}}}+\frac{\color{#3D99F6} \left(-\sin \frac{8\pi }{7}\right)^2}{\sin^2{\frac{4\pi }{7}}} \\ & = \sum_{n=0}^2 \frac{\sin^2\frac{2^{n+1}\pi}7}{\sin^2 \frac{2^n\pi}7} \\ & = \sum_{n=0}^2 \frac{4\sin^2\frac{2^n\pi}7 \cos^2 \frac {2^n\pi}7} {\sin^2 \frac{2^n\pi}7} \\ & = \sum_{n=0}^2 4 \cos^2 \frac {2^n\pi}7 & \small \color{#3D99F6} \text{Note that: } \cos^2 x = \frac 12(\cos 2x +1) \\ & = 2 \sum_{n=0}^2 \left(\cos \frac {2^{n+1}\pi}7 +1\right) \\ & = 2 \left({\color{#3D99F6}\cos \frac {2\pi}7 + \cos \frac {4\pi}7 + \cos \frac {8\pi}7} \right) + 6 & \small \color{#3D99F6} \text{Note that: } \cos (\pi - x) = - \cos x \\ & = {\color{#3D99F6}-} 2 \left({\color{#3D99F6}\cos \frac {5\pi}7 + \cos \frac {3\pi}7 + \cos \frac {\pi}7} \right) + 6 & \small \color{#3D99F6} \text{See note: } \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) = \frac 12 \\ & = - 2 { \color{#3D99F6} \left(\frac 12\right)} + 6 \\ & = \boxed{5} \end{aligned}


Note:

S = k = 0 n 1 cos ( 2 k + 1 2 n + 1 π ) = { k = 0 n 1 e 2 k + 1 2 n + 1 π i } = { e π i 2 n + 1 k = 0 n 1 e 2 k π i 2 n + 1 } = { e π i 2 n + 1 ( 1 e 2 n π i 2 n + 1 1 e 2 π i 2 n + 1 ) } = { e π i 2 n + 1 e π i 1 e 2 π i 2 n + 1 } = { e π i 2 n + 1 + 1 ( 1 + e π i 2 n + 1 ) ( 1 e π i 2 n + 1 ) } = { 1 1 e π i 2 n + 1 } = { 1 1 cos π i 2 n + 1 i sin π i 2 n + 1 } = { 1 cos π i 2 n + 1 + i sin π i 2 n + 1 ( 1 cos π i 2 n + 1 ) 2 + sin 2 π i 2 n + 1 } = { 1 cos π i 2 n + 1 + i sin π i 2 n + 1 2 2 cos π i 2 n + 1 } = 1 2 \small \begin{aligned} S & = \sum_{k=0}^{n-1} \cos \left( \frac {2k+1}{2n+1}\pi \right) \\ & = \Re \left \{ \sum_{k=0}^{n-1} e^{\frac {2k+1}{2n+1}\pi i} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \sum_{k=0}^{n-1} e^{\frac {2k\pi i}{2n+1}} \right \} \\ & = \Re \left \{e^{\frac {\pi i}{2n+1}} \left(\frac {1-e^{\frac {2n\pi i}{2n+1}}}{1-e^{\frac {2\pi i}{2n+1}}}\right) \right \} \\ & = \Re \left \{ \frac {e^{\frac {\pi i}{2n+1}}-e^{\pi i}}{1-e^{\frac {2\pi i}{2n+1}}} \right \} \\ & = \Re \left \{ \frac {e^{\frac {\pi i}{2n+1}} + 1}{\left(1+e^{\frac {\pi i}{2n+1}} \right) \left(1-e^{\frac {\pi i}{2n+1}} \right)} \right \} \\ & = \Re \left \{ \frac 1{1-e^{\frac {\pi i}{2n+1}}} \right \} \\ & = \Re \left \{ \frac 1{1-\cos \frac {\pi i}{2n+1} - i\sin \frac {\pi i}{2n+1}} \right \} \\ & = \Re \left \{ \frac {1-\cos \frac {\pi i}{2n+1} + i\sin \frac {\pi i}{2n+1}}{\left(1-\cos \frac {\pi i}{2n+1}\right)^2 + \sin^2 \frac {\pi i}{2n+1}} \right \} \\ & = \Re \left \{ \frac {1-\cos \frac {\pi i}{2n+1} + i\sin \frac {\pi i}{2n+1}}{2-2\cos \frac {\pi i}{2n+1}} \right \} \\ & = \frac 12 \end{aligned}

@Chew-Seong Cheong Sir , can you please explain me the third last step?

Ankit Kumar Jain - 4 years, 3 months ago

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I have added a note.

Chew-Seong Cheong - 4 years, 3 months ago

Sir , not that part .

The last third = = .

How you converted that 2 n + 1 ? 2^{n + 1}?

Ankit Kumar Jain - 4 years, 3 months ago

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See explanation.

Chew-Seong Cheong - 4 years, 3 months ago

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@Chew-Seong Cheong Thanks.!

Ankit Kumar Jain - 4 years, 3 months ago

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