Trigonometric pattern

Geometry Level 3

n = 1 2018 tan n ! n = 1 2018 tan n ! = ? \large \left \lfloor \ \sum_{n=1}^{2018}\tan{n!^\circ} - \prod_{n=1}^{2018}\tan{n!^\circ} \right \rfloor = \ ?


The answer is -2.

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1 solution

Chew-Seong Cheong
Feb 10, 2018

We note that 6 ! = 720 6! = 720 . This means that n ! n! is a multiple of 180 for n 0 n \ge 0 and t a n n ! = 0 tan n!^\circ = 0 for n 6 n \ge 6 . Therefore,

X = n = 1 2018 tan n ! n = 1 2018 tan n ! = n = 1 5 tan n ! 0 = tan 1 + tan 2 + tan 6 + tan 2 4 + tan 12 0 1.129 \begin{aligned} X & = \sum_{n=1}^{2018} \tan n!^\circ - \prod_{n=1}^{2018} \tan n!^\circ \\ & = \sum_{n=1}^{\color{#D61F06}5} \tan n!^\circ - 0 \\ & = \tan 1^\circ + \tan 2^\circ + \tan 6^\circ + \tan 24^\circ + \tan 120^\circ \\ & \approx - 1.129 \end{aligned}

Therefore, X = 2 \lfloor X \rfloor = \boxed{-2} .

How you approximated the sum without calculator or Programming?

Md Zuhair - 3 years, 4 months ago

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Microsoft Excel spreadsheet.

Chew-Seong Cheong - 3 years, 4 months ago

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Anyway in pen and paper i meant. :)

Md Zuhair - 3 years, 4 months ago

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