6
$\begingroup$

At the end of a calculation it turned out that I wanted to know the value of $$\sin(2\pi/7) + \sin(4\pi/7) - \sin(6\pi/7).$$ Since I knew the answer I was supposed to get, I was able to work out that the the above equals $\sqrt{7}/2$, and I can confirm this numerically. How would I prove this? I suspect I want to use the seventh roots of unity in some way but I am not sure how to proceed.

$\endgroup$
3
  • $\begingroup$ $\sin\bigg(k~\dfrac\pi7\bigg)~$ are the roots of $64x^6-112x^4+56x^2-7$, and $\cos\bigg(k~\dfrac\pi7\bigg)~$ are the roots of $64x^6-~80x^4+24x^2-1$. The latter can be written as $\big(8x^3-4x^2-4x+1\big)\big(8x^3+4x^2-4x-1\big)$, and the cosines corresponding to odd values of k belong to the former, while those corresponding to even values of k belong to the latter. $\endgroup$ – Lucian Oct 16 '15 at 20:05
  • $\begingroup$ @Lucian How do you that the odd values of $k$ belong to the former while even values of $k$ belong to the latter? Is there a reason the polynomial for the sine values does not factor but the polynomial for the cosine values does? And how does knowing these polynomials help me? $\endgroup$ – frakbak Oct 16 '15 at 21:09
  • $\begingroup$ I have revised my solution to a much shorter one using a slightly different approach, but still based on the seventh root of unity. $\endgroup$ – Hypergeometricx Oct 18 '15 at 17:05
2
$\begingroup$

Although it's not very elegant, you can do this calculation by brute force. We have:

$$ \begin{split} i(\sin(2\pi/7)+\sin(4\pi/7)-\sin(6\pi/7)&=\frac{1}{2}(e^{2\pi i/7}-e^{-2\pi i/7}+e^{4\pi i/7}-e^{-4\pi i/7}-e^{6\pi i/7}+e^{-6\pi i/7}) \\ &=\frac{1}{2}(e^{2\pi i/7}-e^{12\pi i/7}+e^{4\pi i/7}-e^{10\pi i/7}-e^{6\pi i/7}+e^{8\pi i/7}) \end{split} $$

squaring this we obtain:

$$ \begin{split} && \frac{1}{4}(e^{2\pi i/7}-e^{12\pi i/7}+e^{4\pi i/7}-e^{10\pi i/7}-e^{6\pi i/7}+e^{8\pi i/7})^2 \\ =&& \frac{1}{4}(\underbrace{e^{4\pi i/7}+e^{10\pi i/7}+e^{8\pi i/7}+e^{6\pi i/7}+e^{12\pi i/7}+e^{2\pi i/7}}_{=-1}-2e^{2\pi i}+2e^{6\pi i/7}-2e^{12\pi i/7}-2e^{8\pi i/7}+2e^{10\pi i/7}-2e^{2\pi i/7}+2e^{8\pi i/7}+2e^{4\pi i/7}-2e^{6\pi i/7}-2e^{2\pi i} -2e^{10\pi i/7}+2e^{12\pi i/7}+2e^{2\pi i/7}-2e^{4\pi i/7}-2e^{2\pi i}) \\=&&\frac{1}{4}(-1-2-2-2)=\frac{-7}{4} \end{split} $$

This gives us:

$$ \begin{split} &&(i(\sin(2\pi/7)+\sin(4\pi/7)-\sin(6\pi/7))^2=\frac{-7}{4} \\ &\implies& -(\sin(2\pi/7)+\sin(4\pi/7)-\sin(6\pi/7))^2=\frac{-7}{4} \\ &\implies& \sin(2\pi/7)+\sin(4\pi/7)-\sin(6\pi/7)=\frac{\sqrt{7}}{2} \end{split} $$

Unfortunately this approach is not very enlightening.

$\endgroup$
1
  • 1
    $\begingroup$ Did you mean $-7/4$ instead of $-7/2$? $\endgroup$ – pregunton Oct 16 '15 at 5:46
3
$\begingroup$

Solution without using trinometry, but only properties of roots of unity:

(This proof has been refined after reviewing this excellent post here. I knew there had to be a more direct approach!:))

Let $\omega=e^{i2\pi/7}$, i.e. the primitive $7$th root of unity. By definition, $$1+\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6=0$$

Let $$\begin{align} S&=\omega^1+\omega^2+\omega^4\\ \Rightarrow\quad S^2&=\omega^2+\omega^4+\omega^8+2(\omega^3+\omega^6+\omega^5)\\ &=2(\underbrace{\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=-1})-(\underbrace{\omega^1+\omega^2+\omega^4}_{=S})\\ &=-2-S\\ S^2+S+2&=0\\ S&=-\frac 12\pm\frac{\sqrt7}2i\\\\ \sin\frac{2\pi}7+\sin\frac{4\pi}7-\sin\frac{6\pi}7 &=\Im(\omega^1+\omega^2-\omega^3)\\ &=\Im(\omega^1+\omega^2+\omega^4)\\ &=\Im(S)\\ &=\frac{\sqrt7}2\quad\blacksquare \end{align}$$

NB - The positive value is chosen as $\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7=\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {\pi}7>0$, since $\sin\frac \pi7<\sin\frac {2\pi}7$.


$\tiny\color{\lightgrey}{\text{Previous solution (now superceded) shown below.}}$ $$\tiny\color{lightgrey}{\begin{align} \sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7 &=\sin\theta+\sin 2\theta-\sin3\theta \qquad\qquad (\theta=2\pi/7;\;\omega=e^{i\theta}=e^{i2\pi/7})\\ &=\frac 1{2i}\left[(\omega^1-\omega^{-1})+(\omega^2-\omega^{-2})-(\omega^3-\omega^{-3})\right]\\ &=\frac 1{2i}\left[(\omega^1-\omega^6)+(\omega^2-\omega^5)-(\omega^3-\omega^4)\right]\\ &=\frac 1{2i}\left[\underbrace{\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=-1}-2(\omega^3+\omega^5+\omega^6)\right]\\ &=-\frac 1{2i}\left[1+2(\omega^3+\omega^5+\omega^6)\right]\\ \left(\sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7\right)^2 &=-\frac14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^3+\omega^5+\omega^6)^2\right]\\ &=-\frac 14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^6+\omega^{10}+\omega^{12}+2\omega^8+2\omega^{11}+2\omega^9)\right]\\ &=-\frac 14\left[1+4(\omega^3+\omega^5+\omega^6)+4(\omega^6+\omega^{3}+\omega^{5}+2\omega^1+2\omega^{3}+2\omega^2)\right]\\ &=-\frac 14\left[8(\underbrace{1+\omega^1+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6}_{=0})-7\right]\\ &=\frac 74\\ \sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7&=\frac{\sqrt7}{\;2}\quad\blacksquare \end{align}}$$

$\endgroup$
2
  • $\begingroup$ @frakbak I think this solution is probably the better one as it provides more insight into what is actually happening. I think really we are looking at the extension $\mathbb{Q}(\zeta_7}/\mathbb{Q}(\sqrt{7})$. The Kronecker-Weber theorem tells us that $\mathbb{Q}(\sqrt{7})$ embeds in a cyclotomic field. I think Gauss sums are the correct tool to use here, which is essentially what is shown above. $\endgroup$ – Sam Weatherhog Oct 18 '15 at 21:50
  • $\begingroup$ @SamWeatherhog - Thank you! Very kind of you. $\endgroup$ – Hypergeometricx Oct 19 '15 at 2:20
1
$\begingroup$

Since $ \sin\left(\frac{6\pi}7\right) = \sin\left(\frac{\pi}7\right) $, then the expression is positive because $ \sin\left(\frac{2\pi}7\right) - \sin\left(\frac{\pi}7\right) > 0 $.

Let $ I $ denote this expression, then $ I > 0 $. Consider

$ I^2 = \left[\underbrace{ \sin^2\left(\frac{2\pi}7\right) + \sin^2\left(\frac{4\pi}7\right) + \sin^2\left(\frac{6\pi}7\right)}_{J} \right] - \left [ \underbrace{2\sin\left(\frac{2\pi}7\right) \sin\left(\frac{4\pi}7\right) - 2\sin\left(\frac{2\pi}7\right) \sin\left(\frac{6\pi}7\right) - 2\sin\left(\frac{4\pi}7\right) \sin\left(\frac{6\pi}7\right)}_{K} \right ] $

By half angle formula, $\sin^2(A) = \frac12 (1-\cos(2A)) $, then $J = \frac32 - \frac12 \left( \cos\left(\frac{4\pi}7\right) + \cos\left(\frac{8\pi}7\right) + \cos\left(\frac{12\pi}7\right) \right) = \frac32 - \frac12\left( \cos\left(\frac{3\pi}7\right) + \cos\left(\frac{5\pi}7\right) + \cos\left(\frac{\pi}7\right)\right) $

$J = \frac32 + \frac12 \times \frac12 = \frac74$

By product to sum formula, $\cos(A) - \cos(B) = -2\sin\left( \frac{A+B}2\right)\sin\left( \frac{A-B}2\right) $, can you show that $ K = 0 $?

$\endgroup$
1
  • $\begingroup$ By roots of unity, it's easy to show that $ \cos\left(\frac{3\pi}7\right) + \cos\left(\frac{5\pi}7\right) + \cos\left(\frac{\pi}7\right) = \frac12 $. Can you show how it's done? $\endgroup$ – GohP.iHan Oct 16 '15 at 2:38
0
$\begingroup$

Using Prosthaphaeresis Formulas,

$$\sin2x+\sin4x-\sin6x=2\sin2x\cos2x-(2\sin2x\cos4x)$$ $$=2\sin2x(\cos2x-\cos4x)=4\sin2x\sin3x\sin x$$

Now from this, $\sin(2n+1)x=(2n+1)\sin x+\cdots+2^{2n}(-1)^n\sin^{2n+1}x$

If $\sin(2n+1)x=0,(2n+1)x=m\pi$ where $m$ is any integer

$x=\dfrac{m\pi}{2n+1}$ where $m\equiv0,\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$

So, the roots of $$2^{2n}(-1)^nt^{2n+1}x+\cdots+(2n+1)t=0$$ are $\sin\dfrac{m\pi}{2n+1}$ where $m\equiv0,\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$

So, the roots of $$2^{2n}(-1)^nt^{2n}x+\cdots+2n+1=0$$ are $\sin\dfrac{m\pi}{2n+1}$ where $m\equiv\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$

$\implies\prod_{r=-n}^n\sin\dfrac{m\pi}{2n+1}=\dfrac{2n+1}{2^{2n}(-1)^n}$

Now as $\sin(-y)=-\sin y,$

$\implies\prod_{r=1}^n(-1)^n\sin^2\dfrac{m\pi}{2n+1}=\dfrac{2n+1}{2^{2n}(-1)^n}$

As $0<\dfrac{m\pi}{2n+1}<\dfrac\pi2$ for $1\le m\le n,$

$\implies\prod_{r=1}^n\sin\dfrac{m\pi}{2n+1}=\dfrac{\sqrt{2n+1}}{2^n}$

Here $n=3$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.