Primal-Trigo summation 2

Geometry Level 4

∑ n = 1 2015 sin ⁡ 5 ( n π 403 ) = ? \large{\sum _{ n=1 }^{ 2015 }{ \sin ^{ 5 }{ \left( \dfrac { n\pi }{ 403 } \right) } } }= \, ?

Give your answer upto 3 decimal places


The answer is 136.830.

Department 8 by Department 8 , 27, Israel

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Otto Bretscher
Oct 25, 2015

By symmetry, we can run the sum from 1 to 403. Now we can use the formulas sin ⁡ 5 ( x ) = 1 16 ( 10 sin ⁡ ( x ) − 5 sin ⁡ ( 3 x ) + sin ⁡ ( 5 x ) ) \sin^5(x)=\frac{1}{16}\left(10\sin(x)-5\sin(3x)+\sin(5x)\right) and ∑ k = 1 n sin ⁡ ( k m π n ) = cot ⁡ ( m π 2 n ) \sum_{k=1}^{n}\sin\left(\frac{km\pi}{n}\right)=\cot\left(\frac{m\pi}{2n}\right) for odd m m to see that the answer is 1 16 ( 10 cot ⁡ ( π 806 ) − 5 cot ⁡ ( 3 π 806 ) + cot ⁡ ( 5 π 806 ) ) ≈ 136.83 \frac{1}{16}\left(10\cot\left(\frac{\pi}{806}\right)-5\cot\left(\frac{3\pi}{806}\right)+\cot\left(\frac{5\pi}{806}\right)\right)\approx \boxed{136.83}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Good usage of those formulas. Expressing sin ⁡ n \sin^n in terms of ( \sin k x ) can be extremely useful in such cases.

Wow, das ist fantastisch! Ich habe noch nie die Formeln bekannt.

Chew-Seong Cheong - 5 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...