Complex Trigono!!

Algebra Level 4

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

Then find the value of f ( 8 ) f\left( 8 \right)


The answer is 9.656.

Mudit Bansal by mudit bansal , 25, India

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1 solution

f ( 8 ) = ∑ r = 1 7 ( 8 − r ) sin ⁡ r π 4 = 7 sin ⁡ π 4 + 6 sin ⁡ π 2 + 5 sin ⁡ 3 π 4 + 4 sin ⁡ π + 3 sin ⁡ 5 π 4 + 2 sin ⁡ 3 π 2 + sin ⁡ 7 π 4 = 7 2 + 6 + 5 2 + 0 − 3 2 − 2 − 1 2 = 4 + 8 2 = 4 + 4 2 = 9.656 \displaystyle \begin{aligned} f(8) & = \sum _{r=1} ^7 {(8-r)\sin {\frac {r\pi}{4}}} \\ & = 7\sin {\frac {\pi}{4}} + 6\sin {\frac {\pi}{2}} + 5\sin {\frac {3\pi}{4}} + 4\sin {\pi} + 3\sin {\frac {5\pi}{4}} + 2\sin {\frac {3\pi}{2}} + \sin {\frac {7\pi}{4}} \\ & = \frac {7}{\sqrt{2}}+6+\frac {5}{\sqrt{2}}+0-\frac {3}{\sqrt{2}}-2-\frac {1}{\sqrt{2}} \\ & = 4 + \frac {8}{\sqrt{2}} \\ & = 4+4\sqrt{2} \\ & = \boxed{9.656} \end{aligned}

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nice solution.Although I was expecting a solution using comlex numbers.I think asking f(8) was a mistake I should have asked f(64) etc.Nonetheless nice solution.

mudit bansal - 6 years, 3 months ago

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Please post a solution using complex numbers . I also did it just by calculating the summation

rohan bansal - 6 years, 3 months ago

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