Prove the given identity.

$$\left(\sin\frac {9\pi}{70}+ \sin\frac {29\pi}{70} - \sin\frac {31\pi}{70}\right) \left(\sin\frac {\pi}{70}-\sin\frac {11\pi}{70} - \sin\frac {19\pi}{70}\right) =\frac {\sqrt {5} -4}{4}$$

Please help, I could not gather enough ideas, even on how to get to the first step. I thought of using the transformation formula (from sum to product) but did not find a fruitful result.


$\left(-\frac{1 +\sqrt{5}}{8} + \frac{1}{8} \sqrt{14\left(5 - \sqrt{5} \right)}\right)\left(-\frac{ 1+ \sqrt{5}}{8} -\frac{1}{8} \sqrt{14 \left(5 -\sqrt{5} \right)}\right) $,

I noticed that Alexis had got to this point. Can anyone explain me how did he get here, with calculations. Please help.

Thanks in Advance.

  • $\begingroup$ Using Taylor expansion, I've rewritten the LHS as $\left[\sum_{n \text{ odd}}^\infty f(n)\left(9^n+29^n-31^n \right)\right]\times\left[\sum_{n \text{ odd}}^\infty f(n)\left( 1-11^n-19^n\right)\right]$ where $f(n)=(-1)^{\left(\frac{n+1}{2}-1\right)}\frac{k^n}{n!}$ with $k=\frac{\pi}{70}$, just in case someone got an idea from this. $\endgroup$ – Bobson Dugnutt Aug 17 '16 at 16:46
  • $\begingroup$ I've noticed that the expression is equivalent to $\left(-\frac{1 + \sqrt{5}}{8} + \frac{1}{8} \sqrt{14 \left(5 - \sqrt{5} \right)}\right) \left(-\frac{ 1+ \sqrt{5}}{8} - \frac{1}{8} \sqrt{14 \left(5 - \sqrt{5} \right)}\right) $, but I'm having difficulty proving this. $\endgroup$ – Alexis Olson Aug 18 '16 at 1:35
  • $\begingroup$ @ Alexis Olson, How did you get to that point? Could you please say me? $\endgroup$ – pi-π Sep 6 '16 at 12:14
  • $\begingroup$ @Lab, I could not relate those hints. Could you please provide more hints? $\endgroup$ – pi-π Sep 10 '16 at 15:36

A method using trigonometric identities, although a very long one -

For simplicity, let me put $\dfrac{\pi}{70}=A$. Now, the expression is -


Multiply out both the brackets completely and for each term of the type "$\sin*\sin$", convert it into sum of two cosines using the formula -

$2\sin{A}\sin{B}=\cos{(A-B)}-\cos{(A+B)}$, to get the following expression -


Now, convert all those cosines whose angles are greater than 90 degrees (i.e. greater than $35A$ here), into cosines of angles less than a 90 degrees (i.e. less than $35A$ here), by using the fact that $\cos{(180-A)}=-\cos{A}$, to get the following simplified expression -

$\dfrac{1}{2}(\cos{8A}-3\cos{10A}-\cos{2A}+3\cos{20A}+3\cos{28A}-3\cos{30A}-\cos{18A}-\cos{22A}+\cos{32A}+\cos{12A})$, which further becomes -







Now use: $\sin{15A}=\cos{20A},\sin{25A}=\cos{10A},\sin{5A}=\cos{30A}$ (Complementary angle pairs, check for yourself!) to write the above expression as -



Now use: $\cos{20A}=-\cos{50A}$ (supplementary angle pairs, check for yourself!) and rewrite the above expression as -

$\dfrac{1}{2}((2\sin{7A}+3)(-\cos{10A}-\cos{30A}-\cos{50A})+3\cos{28A})$ ... (1)

$\cos{10A}+\cos{30A}+\cos{50A}=\dfrac{\cos{30A}\times\sin{30A}}{\sin{10A}}=\dfrac{\sin{60A}}{2\sin{10A}}=\dfrac{1}{2}$. Put this in expression (1) to get -


But $\cos{28A}=\sin{7A}=\dfrac{\sqrt{5}-1}{4}$. Finally put these to get your answer!

  • $\begingroup$ Ahh i was writing the same answer, :( $\endgroup$ – A---B Aug 17 '16 at 19:49
  • $\begingroup$ That's okay...but the point still remains....can we have a better solution? :\ $\endgroup$ – seavoyage Aug 17 '16 at 19:50
  • $\begingroup$ I tried doing it with transformation formulae but it did not get anywhere, we still have to multiple the brackets. $\endgroup$ – A---B Aug 17 '16 at 19:57
  • 2
    $\begingroup$ Wauw, nice (and cumbersome) work! Just some latex tips: Use \times ($\times$) or \cdot ($\cdot$) instead of $*$. Also, to make parenthesis that "fits" what's inside of them, use \left( \right). Cheers :) $\endgroup$ – Bobson Dugnutt Aug 17 '16 at 20:31
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    $\begingroup$ @AlexisOlson: For that, i used the formula for the "sum of cosines of angles in an arithmetic progression".....you can see it here: math.stackexchange.com/questions/117114/… $\endgroup$ – seavoyage Aug 18 '16 at 19:31

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