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Prove the identity $$8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta$$

If $7\theta $ is a multiple of $2\pi,$ Show that $\cos4\theta=\cos3\theta$ and deduce, $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$

My Work

I was able to prove identity using half angle formula and $\cos3\theta $ expansion. Since

$$7\theta=2n\pi$$ $$4\theta=2n\pi-3\theta$$

$$\therefore \cos4\theta=\cos3\theta$$

I cannot prove the final part.

Please help me. Thanks in advance.

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5 Answers 5

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$\cos\frac{0\pi}{7}, \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{6\pi}{7}$ are distinct roots of the fourth order polynomial $$P(x)=8x^4-4x^3-8x^2+3x+1$$

So $P(x)$ can be re-written $$P(x)=8\left(x-\cos\frac{0\pi}{7}\right)\left(x-\cos\frac{2\pi}{7}\right)\left(x-\cos\frac{4\pi}{7}\right)\left(x-\cos\frac{6\pi}{7}\right)$$

Therefore looking at $x^3$ coefficient gives $$\cos\frac{0\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{4}{8}=\frac{1}{2}$$

Actually, for any $n>=2$, $$P_n(x)=T_n(x)-T_{n-1}(x)=2^{n-1}x^n-2^{n-2}x^{n-1}+...$$ Where $T_n$ is the nth Chebyshev polynomial.

So $$P_n(cos(x))=cos(nx)-cos((n-1)x)$$And Likewise, $$\sum_{k=0}^{n-1} cos\frac{2k\pi}{2n-1} = \frac{1}{2}$$

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  • $\begingroup$ Really appreciate it! $x^3$ coefficient will do the trick. $\endgroup$
    – emil
    Commented Nov 10, 2018 at 7:17
  • $\begingroup$ Wait but where did the $P(x) = 8x^4 - 4x^3 -8x^2 + 3x + 1$ come from? $\endgroup$
    – crxyz
    Commented May 27, 2023 at 16:41
  • $\begingroup$ @CrazyVideoGamer See e.g. math.stackexchange.com/questions/4587537/…. $\endgroup$
    – Prasiortle
    Commented Oct 8, 2023 at 9:21
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$$ \begin{align} &1+\color{#C00}{2\cos\left(\frac{2\pi}7\right)}+\color{#090}{2\cos\left(\frac{4\pi}7\right)}+\color{#00F}{2\cos\left(\frac{6\pi}7\right)}\\ &=1+\color{#C00}{\cos\left(\frac{2\pi}7\right)}+\color{#090}{\cos\left(\frac{4\pi}7\right)}+\color{#00F}{\cos\left(\frac{6\pi}7\right)}+\color{#00F}{\cos\left(\frac{8\pi}7\right)}+\color{#090}{\cos\left(\frac{10\pi}7\right)}+\color{#C00}{\cos\left(\frac{12\pi}7\right)}\\ &=\operatorname{Re}\left(1+e^{2\pi i/7}+e^{4\pi i/7}+e^{6\pi i/7}+e^{8\pi i/7}+e^{10\pi i/7}+e^{12\pi i/7}\right)\\ &=\operatorname{Re}\left(\frac{e^{14\pi i/7}-1}{e^{2\pi i/7}-1}\right)\\ &=0 \end{align} $$ Therefore, $$ \cos\left(\frac{2\pi}7\right)+\cos\left(\frac{4\pi}7\right)+\cos\left(\frac{6\pi}7\right)=-\frac12 $$

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  • $\begingroup$ (+1) that's a nice way to do it $\endgroup$
    – TheSimpliFire
    Commented Nov 9, 2018 at 17:18
  • $\begingroup$ This works for $\sum\limits_{k=1}^n\cos\left(\frac{2k\pi}{2n+1}\right)=-\frac12$ $\endgroup$
    – robjohn
    Commented Nov 9, 2018 at 17:59
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Here is one way to do it. It doesn't use your previous work, though.

$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\\ \frac {\sin\frac \pi7(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7})}{\sin \frac \pi7}\\ \frac {\sin\frac \pi7\cos\frac{2\pi}{7}+\sin\frac \pi7\cos\frac{4\pi}{7}+\sin\frac \pi7\cos\frac{6\pi}{7})}{\sin \frac \pi7}\\ \sin A\cos B = \frac12 \sin(A+B) - \frac12\sin(B-A)\\ \frac {\sin\frac {3\pi}7 - \sin \frac {\pi}{7}+\sin\frac {5\pi}7 - \sin\frac{3\pi}{7} +\sin\frac {7\pi}7 - \sin\frac {5\pi}{7}}{2\sin \frac \pi7}\\ \frac {\sin \pi - \sin \frac {\pi}{7}}{2\sin \frac \pi7} = -\frac 12\\ $

Using the info from part 1.

Let $x = \cos \theta$

if $\theta = \frac{2n\pi}{7}$

$8x^4-4x^3 -8x^2+3x+1 = 0$

and $1, \cos \frac{2\pi}{7}, \cos \frac{4\pi}{7}, \cos \frac{6\pi}{7}$ are roots of the polynomial.

$8(x - 1)(x - \cos{2\pi}{7})(x - \cos{4\pi}{7})(x - \cos{6\pi}{7}) = 8x^4-4x^3 -8x^2+3x+1$

By Viteta's rules

$8(1+\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7}) = 4\\ 1+\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7} = \frac 12\\ \cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7} = -\frac 12$

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Since you have obtained sufficient information regarding the route you take to solve the problem, I am giving a different solution. It is a sketch.

Let $\omega:=\exp\left(\dfrac{2\pi\text{i}}{7}\right)$. Show that $\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1=0$. This implies $$\begin{align}\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right)&=\frac{\omega+\omega^{-1}}{2}+\frac{\omega^2+\omega^{-2}}{2}+\frac{\omega^{3}+\omega^{-3}}{2}\\&=\frac{(\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1)-\omega^3}{2\omega^3}=-\frac12\,.\end{align}$$

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  • $\begingroup$ That means you are trying to relate the whole thing to roots of unity, I guess. $\endgroup$
    – Akash Roy
    Commented Nov 9, 2018 at 17:18
  • $\begingroup$ Beautiful answer! $\endgroup$
    – DatBoi
    Commented Jul 19, 2022 at 15:44
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see $ \cos\frac{0\pi}{7}, \cos\frac{2\pi}{7}, \cos\frac{4\pi}{7}, \cos\frac{6\pi}{7}$ are the 4 roots and go by sum of roots formula i.e. **sum of roots = -b/a ** hence, $ \cos\frac{0\pi}{7}+ \cos\frac{2\pi}{7}+ \cos\frac{4\pi}{7}+ \cos\frac{6\pi}{7} = \frac{4}{8}$ $ \cos\frac{2\pi}{7}+ \cos\frac{4\pi}{7}+ \cos\frac{6\pi}{7} =\frac{1}{2} -1 $ =$ \cos\frac{2\pi}{7}+ \cos\frac{4\pi}{7}+ \cos\frac{6\pi}{7} =\frac{-1}{2}$

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