How to prove that $$\left(\sin{\frac{9\pi}{70}}+\sin{\frac{29\pi}{70}}-\sin{\frac{31\pi}{70}}\right)\left(\sin{\frac{\pi}{70}}-\sin{\frac{11\pi}{70}}-\sin{\frac{19\pi}{70}}\right)=\frac{\sqrt{5}-4}{4}?$$ I don't have any idea.
Asked
Active
Viewed 2,254 times
6
-
Not an answer but some inspiration could come from this rewriting : $$\frac{-1-\sqrt{5}+\sqrt{14 (5-\sqrt{5})}}8\times\frac{-1-\sqrt{5}-\sqrt{14 (5-\sqrt{5})})}8$$ – Raymond Manzoni Dec 04 '14 at 12:21
-
use $\sin(\frac{\pi}{10})=\frac{1}{4}(\sqrt{5}-1)$ and rewrite every angel in terms of integer multiples of $\frac{\pi}{10}$ and of $\frac{\pi}{70}$ ... – Math-fun Dec 04 '14 at 12:36
1 Answers
4
Here is an answer, though I'll be amazed if no-one can provide something more elegant. Let $\alpha=e^{i\pi/70}$. Then $\alpha^{70}=-1$ and we have $$LHS=\Bigl(\frac{\alpha^9-\alpha^{-9}}{2i}+etc\Bigr)\Bigl(etc\Bigr)\ ,$$ so $$-4(LHS)=(\alpha^9+\alpha^{61}+etc)(etc)\ .$$ Multiplying everything out on the RHS gives the sum of all $\alpha^k$, where $k$ belongs to the following sets: $$\eqalign{ &\{2,22,42,62,82,102,122\}\cr &\{18,38,58,78,98,118,138\}\cr &\{42,98\}\quad\hbox{twice}\cr &\{10,30,50,90,110,130\}\quad\hbox{three times.}\cr}$$ Now the numbers $\alpha^k$ for the first set are seven points equally spaced around the unit circle, so the sum is zero. Same goes for the second set. Same for the fourth, except that $70$ is missing - no it wasn't a typo! So the sum of $\alpha^k$ for the fourth set is $-\alpha^{70}$, that is, $1$. So we have $$-4(LHS)=2(\alpha^{42}+\alpha^{98})+3=4\cos\frac{3\pi}5+3\ ,$$ and since it is known that $4\cos(3\pi/5)=1-\sqrt5\,$, we are done.
David
- 82,662