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Geometry Level 4

Find the value of ∑ x = 0 7 tan ⁡ 2 π x 16 \displaystyle\sum _{ x=0 }^{ 7 }{ { \tan }^{ 2 }\frac { \pi x }{ 16 } } .


The answer is 35.

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Yash Jain by Yash Jain , 21, India

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2 solutions

∑ x = 0 7 tan ⁡ 2 π x 16 = ∑ x = 1 7 ( sec ⁡ 2 π x 16 − 1 ) = 1 cos ⁡ 2 π 16 + 1 cos ⁡ 2 2 π 16 + 1 cos ⁡ 2 3 π 16 + 1 cos ⁡ 2 4 π 16 + 1 cos ⁡ 2 5 π 16 + 1 cos ⁡ 2 6 π 16 + 1 cos ⁡ 2 7 π 16 − 7 = 1 cos ⁡ 2 π 16 + 1 cos ⁡ 2 2 π 16 + 1 cos ⁡ 2 3 π 16 + 1 cos ⁡ 2 π 4 + 1 sin ⁡ 2 3 π 16 + 1 sin ⁡ 2 2 π 16 + 1 sin ⁡ 2 π 16 − 7 = 1 sin ⁡ 2 π 16 + 1 cos ⁡ 2 π 16 + 1 sin ⁡ 2 2 π 16 + 1 cos ⁡ 2 2 π 16 + 1 sin ⁡ 2 3 π 16 + 1 cos ⁡ 2 3 π 16 + 2 − 7 = 1 sin ⁡ 2 π 16 cos ⁡ 2 π 16 + 1 sin ⁡ 2 2 π 16 cos ⁡ 2 2 π 16 + 1 sin ⁡ 2 3 π 16 cos ⁡ 2 3 π 16 − 5 = 4 sin ⁡ 2 π 8 + 4 sin ⁡ 2 π 4 + 4 sin ⁡ 2 3 π 8 − 5 = 4 sin ⁡ 2 π 8 + 4 cos ⁡ 2 π 8 + 8 − 5 = 16 sin ⁡ 2 π 4 + 3 = 32 + 3 = 35 \begin{aligned} \sum_{x=0}^7 \tan^2{\frac{\pi x}{16}} & = \sum_{x=\color{#D61F06}{1}}^7 \left( \sec^2{\frac{\pi x}{16}} - 1\right) \\ & = \frac{1}{\cos^2{\frac{\pi}{16}}} + \frac{1}{\cos^2{\frac{2\pi}{16}}} + \frac{1}{\cos^2{\frac{3\pi}{16}}} + \frac{1}{\cos^2{\color{#D61F06}{\frac{4\pi}{16}}}} \\ & \quad \quad + \color{#3D99F6}{\frac{1}{\cos^2{\frac{5\pi}{16}}} + \frac{1}{\cos^2{\frac{6\pi}{16}}} + \frac{1}{\cos^2{\frac{7\pi}{16}}}} - 7 \\ & = \frac{1}{\cos^2{\frac{\pi}{16}}} + \frac{1}{\cos^2{\frac{2\pi}{16}}} + \frac{1}{\cos^2{\frac{3\pi}{16}}} + \frac{1}{\cos^2{\color{#D61F06}{\frac{\pi}{4}}}} \\ & \quad \quad + \color{#3D99F6}{\frac{1}{\sin^2{\frac{3\pi}{16}}} + \frac{1}{\sin^2{\frac{2\pi}{16}}} + \frac{1}{\sin^2{\frac{\pi}{16}}}} - 7 \\ & = \frac{1}{\sin^2{\frac{\pi}{16}}} + \frac{1}{\cos^2{\frac{\pi}{16}}} + \frac{1}{\sin^2{\frac{2\pi}{16}}} + \frac{1}{\cos^2{\frac{2\pi}{16}}} \\ & \quad \quad + \frac{1}{\sin^2{\frac{3\pi}{16}}} + \frac{1}{\cos^2{\frac{3\pi}{16}}} + \color{#D61F06}{2} - 7 \\ & = \frac{1}{\sin^2{\frac{\pi}{16}} \cos^2{\frac{\pi}{16}}} + \frac{1}{\sin^2{\frac{2\pi}{16}} \cos^2{\frac{2\pi}{16}}} + \frac{1}{\sin^2{\frac{3\pi}{16}} \cos^2{\frac{3\pi}{16}}} -5 \\ & = \color{#3D99F6}{\frac{4}{\sin^2{\frac{\pi}{8}}}} + \color{#D61F06}{ \frac{4}{\sin^2{\frac{\pi}{4}}}} + \color{#3D99F6} {\frac{4}{\sin^2{\frac{3\pi}{8}}}} -5 \\ & = \color{#3D99F6}{\frac{4}{\sin^2{\frac{\pi}{8}}} + \frac{4}{\cos^2{\frac{\pi}{8}}}} + \color{#D61F06}{8} - 5 = \color{#3D99F6}{\frac{16}{\sin^2{\frac{\pi}{4}}}} + 3 = \color{#3D99F6}{32} + 3 = \boxed{35} \end{aligned}

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Moderator note:

There's a slightly shorter approach to this problem.

Hint : Prove that 2 cot ⁡ ( 2 x ) = tan ⁡ ( x ) − cot ⁡ ( x ) 2\cot(2x) = \tan(x) - \cot(x) .

Ronak Agarwal
Jul 4, 2015

Check this

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Moderator note:

Thanks for sharing the more general form of

∑ r = 1 2 n − 1 − 1 tan ⁡ 2 ( r π 2 n ) \sum _{ r=1 }^{ { 2 }^{ n-1 }-1 } { \tan ^{ 2 } { \left (\frac { r\pi }{ { 2 }^{ n } } \right ) } }

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