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Prove that $\cos\theta$ is root of equation $8x^3-4x^2-4x+1=0$, given $\theta=\frac{\pi}{7}$.

I put $\cos\theta$ in equation, but couldn't show the left-hand side to be zero.

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    $\begingroup$ $\cos \frac{\pi}{7}$ is not a root of this equation as you can see its roots in cubic equation calculator, for example 1728.org/cubic.htm $\endgroup$
    – Arun
    Feb 11, 2017 at 6:12
  • $\begingroup$ @dxiv yes edited thanks $\endgroup$
    – Gathdi
    Feb 11, 2017 at 6:20
  • $\begingroup$ @ClaudeLeibovici thanks a lot $\endgroup$
    – Gathdi
    Feb 11, 2017 at 6:28

4 Answers 4

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Put $2x=z+\dfrac1z$ in $$(2x)^3-(2x)^2-2(2x)+1=0$$

and multiply by $z+1$ to find $z^7+1=0$ whose roots are are $e^{(2k+1)\pi i/7}$ where $k\equiv0,\pm1,\pm2,\pm3\pmod7$

So, the roots of $$\dfrac{z^7+1}{z+1}=0$$ are $e^{(2k+1)\pi i/7}$ where $k\equiv0,\pm1,\pm2,3\pmod7$

Finally $2\cos y=e^{iy}+e^{-iy}$

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$$ \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}= $$ $$ =\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2}. $$ Let $\cos\frac{\pi}{7}=x$.

Hence, $$2x^2-(4x^3-3x)-x=-\frac{1}{2},$$ which gives your equation.

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  • $\begingroup$ how did you write first step?i.e $cos2\theta+cos4\theta +$ $\endgroup$
    – Gathdi
    Feb 11, 2017 at 7:07
  • $\begingroup$ @Gathdi I knew that it shrinks. $\endgroup$ Feb 11, 2017 at 7:25
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We have $$ \cos{2\theta} = 2\cos^2{\theta}-1 \\ \cos{3\theta} = \cos{\theta}(2\cos^2{\theta}-1) - 2\sin^2{\theta}\cos{\theta} = 4\cos^3{\theta}-3\cos{\theta}, $$ so $$ \cos^2{\theta} = \frac{1}{2}(\cos{2\theta}+1) \\ \cos^3{\theta} = \frac{1}{4}(\cos{3\theta}+3\cos{\theta}) $$

Putting these into the equation gives $$ 8\cos^{3}{\theta}-4\cos^2{\theta}-4\cos{\theta}+1 \\ = 2\cos{3\theta}+6\cos{\theta} -2\cos{2\theta}-2-4\cos{\theta}+1 \\ = -1+2(\cos{3\theta}-\cos{2\theta}+\cos{\theta}) = \frac{\cos{(7x/2)}}{\cos{(x/2)}}, $$ the last part of which comes from the formula $$ \sum_{k=-n}^n (-1)^k \cos{kx} = (-1)^n \frac{\cos{(n+1/2)x}}{\cos{(x/2)}}, $$ which can be proven by induction. It's then clear that this is zero if $x$ is a zero of $\cos{(7x/2)}$, but not $\cos{(x/2)}$, and the first one of these is $\theta=\pi/7$.

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  • $\begingroup$ i didnot understand your last two lines of solution $\endgroup$
    – Gathdi
    Feb 11, 2017 at 7:09
  • $\begingroup$ The $\cos{(7x/2)}=0$ part, or something else? $\endgroup$
    – Chappers
    Feb 11, 2017 at 13:00
  • $\begingroup$ Yeah that one.. $\endgroup$
    – Gathdi
    Feb 12, 2017 at 4:07
  • $\begingroup$ If $x=\cos{\theta}$, then I have shown that $8x^3-4x^2-4x+1 = \frac{\cos{(7x/2)}}{\cos{(x/2)}}$. Therefore the LHS is zero if and only if the right hand side is. This happens if the numerator is zero, but the denominator is not (because if both are zero the fraction may be $0$, $\infty$ or something else; it turns out that in this case, the zeros of both are simple, so it's not zero). Now, what we are actually interested in is that if $\theta=\pi/7$, $\cos{(7x/2)} = \cos{(\pi/2)} = 0$, while $\cos{(\pi/14)} \neq 0$, so the RHS is zero as required. $\endgroup$
    – Chappers
    Feb 12, 2017 at 19:03
  • $\begingroup$ You mentioned some formula in last part. Can you provide link to its proof? Also can we get that fromula if we apply CosC+cosD formulas? $\endgroup$
    – Gathdi
    Feb 13, 2017 at 4:06
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We can use theory of equations (and some disguised Galois theory).

Given equation can be rewritten as $(2x)^3-(2x)^2 -2( 2x)+1=0$. So we need to show $2\cos \theta $ is a root of $x^3-x^2-2x+1=0$. We will do this by showing there is a cubic equation with integer coefficients satisfied by $2\cos\theta$, and determine the other two roots, and from that reconstruct the equation.

First let us write $\pi/7 = 2\pi/14=\theta$. Let $\alpha=e^{2\pi i/14}$, a primitive 14th root of unity.

Clearly $-\alpha$ is a primitive $7$th root of unity. The latter is any solution of $x^6+x^5+x^4+x^3+x^2+x+1=0$. This tells us that $\alpha$ is a root of $x^6-x^5+x^4-x^3+x^2-x+1=0$.

Now $\alpha+\bar\alpha = 2\cos\theta=\alpha+\alpha^{13}$. We will compute the polynomials satisfied by this number.

The Galois conjugates (the other roots) of $\alpha+\alpha^{13}$ are $\alpha^3+\alpha^{11},\alpha^5+\alpha^9$ (pair numbers less than 14 and coprime to it such that they add up to 14, and sum the corresponding powers of $\alpha$.)

It is a cubic with roots $a,b,c$ where $a=\alpha+\alpha^{13}, b=\alpha^3+\alpha^{11}, c=\alpha^5+\alpha^9$. So we need to calculate $a+b+c, ab+bc+ca$ and $abc$. So it boils down to showing $a+b+c=1, ab+bc+ca=-1, abc=-1$. Now this is a routine verification using the fact that $\alpha^{14}=1$,

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