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.

  • $\begingroup$ $\sin\bigg(k~\dfrac\pi7\bigg)~$ 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

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.

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

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 $\sin\frac \pi7<\sin\frac {2\pi}7$.

$\tiny\color{\lightgrey}{\text{Previous solution (now superceded) shown below.}}$ $$\tiny\color{lightgrey}{\begin{align} \sin\frac {2\pi}7+\sin\frac {4\pi}7-\sin\frac {6\pi}7 &=\sin\theta+\sin 2\theta-\sin3\theta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

  • $\begingroup$ @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

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 $?

  • $\begingroup$ 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

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 $m\equiv\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$


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


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


Here $n=3$


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